Induction Proof: $(1+x)^n = 1+x^n$ for even $n$ in $\mathbb{F}_2[x]$ I'm trying to work on a proof by induction. The statement is: 
Let n be even. Then, $(x+1)^n = x^n + 1$ for $n\in\mathbb{N}\cup{0}$ and $(x^n+1) \in  \mathbb{F}_2[x]$
Base case: $n=0$ and $n=2$ both satisfy the condition, fairly trivially. I include the $n=2$ case specifically to draw upon later. 
We suppose true for $n=2k$ (as we require n even), so that we assume:
$(x+1)^{2k} = x^{2k} + 1$
Then, if $n=2k+2$:
$(x+1)^{2k+2} = (x+1)^{2k} (x+1)^2$ 
$\Rightarrow (x+1)^{2k+2} = (x^{2k}+1)(x^2+1)$, where the first RHS bracket follows from the inductive assumption, and the second bracket was shown to hold earlier. 
Then, expanding RHS gives: 
$(x+1)^{2k+2} = x^{2k+2} + 1 + x^2 + x^{2k}$
This is where I come unstuck. I'm hoping to show that $(x+1)^{2k+2} = x^{2k+2} + 1$ , but I can't see a way to get rid of the extra terms on the RHS to complete my proof. 
For what it's worth, the actual question I'm working on simply concerns the case $n=6$, and while I could do this by showing that (x+1) is a factor, then divide it out, and then show that (x+1) is still a factor, and repeat 6 times, that feels like an unusually ugly approach. I can't see any other obvious way to handle the problem, so hints appreciated.

What you are trying to prove isn't true: $(1 + x)^n = 1 + x^n$ in $\Bbb{F_2}[x]$ iff $n$ is a power of two (check the case $n=6$ if you don't believe me). Sierpiniski's triangle gives a nice visualisation of how the binomial coefficients work modulo $2$.
