# $a \equiv b$ (mod $p$) implies $a^{p^n} \equiv b^{p^n}$ (mod $p^n$)?

Let $$p$$ be a prime number. If $$a \equiv b$$ (mod $$p$$), does that imply $$a^{p^n} \equiv b^{p^n}$$ (mod $$p^n$$)?

I think the answer will be yes, and I suspect that the way of proving it will involve writing $$a^{p^n}-b^{p^n}$$ as a multiple of $$(a-b)^n$$.

I also noticed that $$n, and I'm wondering if this will make the proof easier or not.

$$a=b+kp$$ where $$k$$ is an integer
$$(b+kp)^{p^n}=b^{p^n}+\binom{p^n}1b^{p^n-1}kp+\binom{p^n}2b^{p^n-2}(kp)^{2}+\cdots+(kp)^{p^n}$$
$$\equiv b^{p^n}\pmod{p^{n+1}}$$