Fermat's Last Theorem; A Simple Proof 2017

Things don't get much simpler than this but it still took me the best part of seven years to see it!
Edited Nov 2021 : And after a further 4 years I can now see that it is an absolute proof because Z^(2n+1) for Z=(A+B) an integer can ALWAYS be expanded algebraically by the binomial theorem for all the (A+B) binary partitions of Z.
Likewise all X^(2n+1)+(Y^2n+1) are divisible by (X+Y) to give two factors (X+Y)( algebraic expansion) due to the division. These are also very predictable in form so what we see for small numbers also applies to infinitely large ones.
Now the binomial expansion applicable to one side of the Fermat equation can never equal the binomial number expansion that applies to the other side of the Fermat equation since they are both totally different hence the Fermat equation is FALSE and Pierre de Fermat was right and the means to prove it in his era were available. Q.E.D.

[MuffinsAPlenty]

How does step 3 follow from step 2? I have trouble seeing how (x+y) dividing z^(2n+1) implies that (x+y)^(2n+1) does too.

[mambu]

The remedy you suggest in your description for step 3 is not correct you try to prove (x+y) divides z^(2n+1) implies (x+y)^(2n+1) divides z^(2n+1) by using the fact that x+y divides all odd powers of z
But this is not true you only know that (x+y) divides z^(2n+1) for that FIXED Odd number (2n+1)

[noxicate]

Alastair Bateman what is your email! I need your help!