Fermat's Last Theorem, here's a broader margin.

So, about Fermat's Last Theorem where a,b,c are integers

${ a }^{ n }+{ b }^{ n }={ c }^{ n }\quad Let\quad x+1=a,\quad y+1=b,\quad z=c,$ so $a,b>1$

We have

${ \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }-{ z }^{ n }=0$

Consider

${ \left[ \left( x+1 \right) +\left( y+1 \right)\right]}^{ n }=\quad { \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ 1 \end{matrix} \right) \left( x+1 \right) { \left( y+1 \right)}^{ n-1 }\\\quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ 2 \end{matrix} \right) { \left( x+1 \right)}^{ 2 }{ \left( y+1 \right)}^{ n-1 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +...\\\quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ n-1 \end{matrix} \right) { \left( x+1 \right)}^{ n-1 }{ \left( y+1 \right)}^{ 1 }$

then

${ \left[ \left( x+1 \right) +\left( y+1 \right)\right]}^{ n }-{ z }^{ n }>{\left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }-{ z }^{ n }$

Also,

${ \left( x+1 \right)}^{ n }={ x }^{ n }+\left( \begin{matrix} n \\ 1 \end{matrix} \right) { x }^{ n-1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { x }^{ n-2 }+...+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) x+1\\ { \left( y+1 \right)}^{ n }=y^{ n }+\left( \begin{matrix} n \\ 1 \end{matrix} \right) y^{ n-1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { y }^{ n-2 }+...+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) y+1\\ { \left( z+1 \right) }^{ n }=z^{ n }+\left( \begin{matrix} n \\ 1 \end{matrix} \right) z^{ n-1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { z }^{ n-2 }+...+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) z+1$

We have,

${ \left[ \left( x+1 \right) +\left( y+1 \right)\right]}^{ n }-{ z }^{ n }=\quad { x }^{ n }+y^{ n }-{ z }^{ n }\\\quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ 1 \end{matrix} \right) \left[ { x }^{ n-1 }+\left( x+1 \right) { \left( y+1 \right)}^{ n-1 }+{ y }^{ n-1 } \right] \\\quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ 2 \end{matrix} \right) \left[ { x }^{ n-2 }+{ \left( x+1 \right)}^{ 2 }{ \left( y+1 \right)}^{ n-2 }+{ y }^{ n-2 } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +...\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ n-1 \end{matrix} \right) \left[ x+{ \left( x+1 \right)}^{ n-1 }\left( y+1 \right) +y \right] \\\quad \quad \quad \quad \quad \quad \quad \quad \quad +2$

${ \left[ \left( x+1 \right) +\left( y+1 \right)\right]}^{ n }-{ z }^{ n }=\quad { x }^{ n }+y^{ n }-{ z }^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ 1 \end{matrix} \right) \left[ { x }^{ n-1 }+{ f }_{ 1 }\left( x,y \right) +{ y }^{ n-1 } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ 2 \end{matrix} \right) \left[ { x }^{ n-2 }+{ f }_{ 2 }\left( x,y \right) +{ y }^{ n-2 } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +...\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +\left( \begin{matrix} n \\ n-1 \end{matrix} \right) \left[ x+{ f }_{ n-1 }\left( x,y \right) +y \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +2\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad +\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)}$

then

${ \left[ \left( x+1 \right) +\left( y+1 \right)\right]}^{ n }-{ z }^{ n }>{ \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }-{ z }^{ n }$


${ \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }-{ z }^{ n }>{ x }^{ n }+{ y }^{ n }-{ z }^{ n }+2+\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)}$-----------(2a)


${ \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }-{ z }^{ n }>{ x }^{ n }+{ y }^{ n }-{ z }^{ n }$ ---------------------------(1)


${ x }^{ n }+y^{ n }-{ z }^{ n }+2+\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)}>{ x }^{ n }+y^{ n }-{ \left( z+1 \right) }^{ n }+2+\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)}$ -----------(2b)

and so, when

${ \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }-{ z }^{ n }=0$ -------------(S)

then from expression (2a) and (2b)

${ x }^{ n }+y^{ n }-{ \left( z+1 \right) }^{ n }+2+\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)}<0$

and from expression (1)

${ x }^{ n }+{ y }^{ n }-{ z }^{ n }<0$

that is ${ \left( z+1 \right)}^{ n }-{ z }^{ n }<2+\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)}$

${ \left( z+1 \right)}^{ n }-{ z }^{ n }=\left( \begin{matrix} n \\ 1 \end{matrix} \right) z^{ n-1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { z }^{ n-2 }+...+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) z+1$

         $=\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)z^{ n-i }+1}<2+\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)}$  -------------(3)

for $n=2$

$\left( \begin{matrix} 2 \\ 1 \end{matrix} \right) z+1<\sum _{ 1 }^{ 1 }{ \left( \begin{matrix} 2 \\ i \end{matrix} \right)} \quad 2z<3$ which can be true. (This is a statment that was derived from the starting condition (S) that prove to be possible.)

but for $n=3$

$\left( \begin{matrix} 3 \\ 1 \end{matrix} \right) z^{ 2 }+\left( \begin{matrix} 3 \\ 2 \end{matrix} \right) z+1<2+\sum _{ i=1 }^{ 2 }{ \left( \begin{matrix} 3 \\ i \end{matrix} \right)}$

$3{ z }^{ 2 }+3z\nless7$ for all $z > 1$.

In fact we have a contradiction for all $n>2$ because the expression $\sum _{ i=1 }^{ n-1 }{ \left( \begin{matrix} n \\ i \end{matrix} \right)}$ is just the coefficients of the $z$ on the L.H.S. of  expression (3).  (For $n>2$ the derived statment proved to be false, invalidating the starting condition (S).) And so, ${ \left( x+1 \right)}^{ n }+{ \left( y+1 \right)}^{ n }\neq{ z }^{ n }$ for $n>2$.

Therefore,

${ a }^{ n }+{ b }^{ n }={ c }^{ n }$ has no valid solution for $n>2$ and $a,b>1$

And that's Fermat's Last Theorem, proved by contradiction.
(Expression (3) is derived from (S), it is not an attempt to solve for $z$.)