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.

Note, the last digit of $n^5$ can only be affected by the last digit of $n$.  That is, the 1s digit of any power of $n$ will only be affected by the ones digit of $n$, and not the 10s, 100s, or any other digit.  So, you only have 10 cases, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.  All you have to do is try them separately.  $0^5 = 0$,  $1^5 = 1$, $2^5 = 32$, $3^5 = 243$, ..., $9^5 = 59049$.
If you've heard of modular arithmetic, all numbers mod 10 are going to be in the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.  And, this is why we only need consider these 10 cases.
