Proof that $6^n$ always has a last digit of $6$ Without being proficient in math at all, I have figured out, by looking at series of numbers, that $6$ in the $n$-th power always seems to end with the digit $6$.
Anyone here willing to link me to a proof?
I've been searching google, without luck, probably because I used the wrong keywords.

HINT $\rm\ \ 6-1\ |\ 6^k-1,\ $ so $\rm\:\ 2,5\ |\ 6^n-6\ \Rightarrow\ 10\ |\ 6^n - 6\:,\ $ i.e. $\rm\ 6^n\ =\ 6 + 10\ k\:$ for $\rm\:k\in\mathbb Z\:.$
Alternatively: $\rm\ mod\ 10:\ \ 6^n\equiv 6\ $ since it is $\rm\ 0^n \equiv 0\pmod 2,\ \ 1^n \equiv 1\pmod 5$
Similarly odd $\rm\:b\: \Rightarrow\: (b+1)^n\equiv b+1\pmod{2\:b}\:,\:$ so $\rm\:(b+1)^n\:$ has last digit $\rm\:b+1\:$ in radix $\rm\:2\:b\:.$
NOTE  how modular arithmetic reduces the induction to the trivial inductions $\rm\ 0^n = 0,\ 1^n = 1\:.$ This is a prototypical example of the sort of simplification afforded by reducing arithmetical problems to their counterparts in the simpler arithmetical rings of integers $\rm\:(mod\ m)\:.\:$
