modular arithmetic - Show $a^k ≡ b^kpmod m$

I need to show that if $a,b,k$ and $m$ are integers and $k ≥ 1, m ≥ 2$, and $a ≡ b\pmod m$, then: $a^k ≡ b^k \pmod m$.

But I have no idea how to show this, I have never been this confused. So is there anyone who could help? just a little, like I honestly feel overwhelmed (sounds stupid I know, sorry)

*what do i need to do with the m ≥ 2 ???

Answer

I assume you know (all equivalences are $\text{mod } m$)

1. $a\equiv b \iff a-b\equiv 0$


2. $c\equiv 0 \implies cd\equiv 0$

Then

$\begin{align}&a\equiv b\\ \iff &a-b\equiv 0\\ \implies &(a-b)(a^{k-1} + a^{k-2}b + \cdots + ab^{k-2}+b^{k-1})\equiv 0\\ \iff &a^k-b^k\equiv 0\\ \end{align}$

The last deduction above is technically hand-waved but can be made formal with a summation or with induction.