Inductive step in proof of Freshman's Dream I am trying to prove that for $K$ a field of characteristic $p$ prime, $q$ a power of $p$ and $x,y$ in $K$,
$$(x+y)^q=(x^q + y^q).$$
I have the base case, and now I am trying to do the inductive step.
So far I have that 
$$(x+y)^{p^{n+1}}=(x+y)^{p^n}(x+y)^p=(x^{p^n}+y^{p^n})(x^p+y^p)=x^{p^{n+1}}+y^{p^{n+1}}+y^px^{p^n}+y^{p^n}x^p.$$
So it just follows that I need to prove that $y^px^{p^n}+y^{p^n}x^p=0.$
I read that $x^p=1$ and $x^{p^n}=x$, so this would imply that $y^px^{p^n}+y^{p^n}x^p=x+y$ but this does not help fulfil the theorem.
What is the step I am missing?
 A: You have a mistake in your first equality:
$(x+y)^{p^{n+1}} = (x+y)^{p^n \cdot p}$, while $(x+y)^{p^n}(x+y)^p = (x+y)^{p^n+p} \neq (x+y)^{p^{n+1}}$.  Correcting this mistake fixes everything:
\begin{align*}
(x+y)^{p^{n+1}} &= (x+y)^{p^n \cdot p} = ((x+y)^{p^n})^p = (x^{p^n} + y^{p^n})^p = (x^{p^n})^p + (y^{p^n})^p = x^{p^{n+1}} + y^{p^{n+1}} \, .
\end{align*}
A: So your base case is $(x+y)^p=x^p+y^p$. So if we assume the case $q=p^k$, then we have
$$
(x+y)^{p^{k+1}}=\left((x+y)^{p^k}\right)^p.
$$
Our inductive hypothesis tells us that the inside is
$$
(x+y)^{p^k}=x^{p^k}+y^{p^k}.
$$
Our base case tells us that
$$
\left((x+y)^{p^k}\right)^p=\left(x^{p^k}+y^{p^k}\right)^p=\left(x^{p^k}\right)^p+\left(y^{p^k}\right)^p=x^{p^{k+1}}+y^{p^{k+1}}.
$$
A: You can prove it without induction: you can see $x\longmapsto x^p$ as a ring morphism (the Frobenius morphism, $\operatorname{Fr}$. Then $x\longmapsto x^{p^n}$ is just the iterated $\operatorname{Fr}^n$ and the composition of ring morphisms is a ring morphism.
Of course there is an implicit induction behind this argument, but it's so trivial we don't have to write it.
