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.