Prime Power of Sum Modulo Prime

Theorem
Let $p$ be a prime number.

Then:
 * $\forall n \in \N_{> 0}: \paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
 * $\paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$

Basis for the Induction
First from Power of Sum Modulo Prime we have that $\map P 1$ is true:
 * $\paren {a + b}^p \equiv a^p + b^p \pmod p$

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\paren {a + b}^{p^k} \equiv a^{p^k} + b^{p^k} \pmod p$

Then we need to show:
 * $\paren {a + b}^{p^{k + 1} } \equiv a^{p^{k + 1} } + b^{p^{k + 1} } \pmod p$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N_{> 0}: \paren {a + b}^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$

Also see

 * Power of Sum Modulo Prime