How to prove that every power of 6 ends in 6? Yesterday I had the traditional math matriculation exam, and in it there was a question "In what digit does the number $2016^{2016}$ end in?" After the test The Matriculation Examination Board published a pdf in which they show how to basically solve all the problems in the test, and for the aforementioned question the solution was "Because every power of 6 ends in 6, so does every power of 2016, and thus the last digit is 6". I understand everything else in that problem now, except how to actually show that every power of 6 ends in 6. From there on I know how to solve the last digit if $2016^{2016}$. So,

how to prove that every power of 6 ends in 6?


Use induction in order to complete the hint given by @ThePortakal.

First, show that this is true for $n=1$:
$6^1=6$
Second, assume that this is true for $n$:
$6^n=10k+6$
Third, prove that this is true for $n+1$:
$6^{n+1}=$
$6\cdot\color{red}{6^n}=$
$6\cdot(\color{red}{10k+6})=$
$60k+36=$
$60k+30+6=$
$10(6k+3)+6$

Please note that the assumption is used only in the part marked red.
