Use induction to prove that $(a+b)^{p^n}=a^{p^n}+b^{p^n}$

Question

Use induction to prove that $(a+b)^{p^n}=a^{p^n}+b^{p^n}$ where $a,b \in F$ and $char(F)=p$.

So, i'm having problems doing this like a normal induction problem because i'm doing induction on the exponent of an exponent so the algebra gets messy real quick, so basically i'm having trouble getting the induction hypothesis involved during the inductive step. Should I write this as a sum with the binomial expansion and then break up the summation into two parts, one being from $0$ to $p^n$ and the other being from $p^n+1$ to $p^{n+1}$? Is there a better way to do this? Thanks in advance!

Answer 1

$$ (a+b)^{p^0} = (a+b)^1 = a + b = a^{p^0} + b^{p^0}\\ $$

Suppose by induction $(a+b)^{p^n}=a^{p^n}+b^{p^n}$

$$ (a+b)^{p^{n+1}} = (a+b)^{p^n p} = ((a+b)^{p^n})^p\\ = (a^{p^n} + b^{p^n})^p\\ = a^{p^{n+1}} + b^{p^{n+1}} $$

where in the last line we use binomial formula in characteristic $p$

$$ (x+y)^p = \sum \binom{p}{k} x^k y^{p-k} $$

All terms except $\binom{p}{p} x^k$ and $\binom{p}{0}y^p$ vanish because then $\binom{p}{k}$ is divisible by p.