Prime Power of Sum Modulo Prime

Theorem
Let $p$ be a prime number.

Then:
 * $\forall n \in \N^*: \left({a + b}\right)^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$

Corollary
Let $p$ be a prime number.

Then:
 * $\forall n \in \N^*: \left({1 + b}\right)^{p^n} \equiv 1 + b^{p^n} \pmod p$

Proof
Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\left({a + b}\right)^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$

Basis for the Induction
First we need to show that $P(1)$ is true:
 * $\left({a + b}\right)^p \equiv a^p + b^p \pmod p$

From the Binomial Theorem: $\left({a + b}\right)^p = \sum_{k=0}^p \binom p k a^k b^{p-k}$.

Also note that $\sum_{k=0}^p \binom p k a^k b^{p-k} = a^p + \sum_{k=1}^{p-1} \binom p k a^k b^{p-k} + b^p$.

So:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

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

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

Induction Step
This is our induction step:

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

Therefore:
 * $\forall n \in \N^*: \left({a + b}\right)^{p^n} \equiv a^{p^n} + b^{p^n} \pmod p$

Proof of Corollary
Follows immediately by putting $a=1$.

Also see
Compare with the Freshman's Dream.