Why is the last digit of $n^5$ equal to the last digit of $n$? I was wondering why the last digit of $n^5$ is that of $n$? What's the proof and logic behind the statement? I have no idea where to start. Can someone please provide a simple proof or some general ideas about how I can figure out the proof myself? Thanks.

The claim is clearly equivalent to the claim that $n^5 - n$ is divisible by $10$.
By Fermat's little theorem, $n^5 - n$ is divisible by $5$.
$n^5 - n = n(n^4 - 1)$ is even because if $n$ is odd, then $n^4-1$ is even.
Thus, $n^5 - n$ is divisible by both $5$ and $2$, and, these being coprime, it follows that $n^5 - n$ is divisible by $10$.
