$a^k\equiv b^k\bmod n$ by induction

Question

$$a,b\in\Bbb Z$$ $$n,k\in\Bbb N^+$$ $$a\equiv b\bmod n$$ $$\text{Use induction to prove}$$ $$a^k\equiv b^k\bmod n$$

Answer 1

The base case is a given.

The successor step follows from

$$x \equiv y \pmod{n} \land x' \equiv y' \pmod{n} \implies xx' \equiv yy' \pmod{n}$$.

Answer 2

Use the induction hypothesis to suppose $a^k\equiv b^k\mod n$. Using the fact that: if $a\equiv b\mod n$ and $c\equiv d\mod n$ then $ac\equiv bd\mod n$ we have: $$\begin{align}aa^k&\equiv bb^k\mod n\qquad\text{since $a\equiv b\mod n$}\\\implies a^{k+1}&\equiv b^{k+1}\mod n\end{align}$$