Let $n$ be a natural number. Prove that $6^n ≡ 6 \pmod {10}$.
What conclusion can you draw about the last digit of powers of 6?
We'll prove this by induction.
Let the statement $P(n)$ mean that $6^n \equiv 6 \pmod {10}$.
We need to prove the base case $P(1)$, and the inductive step $P(n) \implies P(n+1)$.
Base Case
The base case $P(1)$ is that $6^1 \equiv 6 \pmod {10}$. This is trivially true.
Induction Step
We assume $P(n)$ and aim to show $P(n+1)$.
$P(n+1)$ is
$$ \begin{align} 6^{n+1} & \equiv 6^n \times 6 \pmod{10} \\ \\ & \equiv 6 \times 6 \pmod {10} \tag{*} \\ \\ & \equiv 36 \pmod {10} \\ \\ & \equiv 6 \pmod {10} \end{align}$$
Line (*) uses the induction hypothesis that $6^n \equiv 6 \pmod {10}$.
We have shown that $P(n) \implies P(n+1)$.
We have shown, by induction, that $6^n \equiv 6 \pmod {10}$. We conclude that the last digit of powers of 6 is always 6.