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

Question

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?

Answer 1

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

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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 $n+1$:

$6^{n+1}=$

$6\cdot\color{red}{6^n}=$

$6\cdot(\color{red}{10k+6})=$

$60k+36=$

$60k+30+6=$

$10(6k+3)+6$

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

Answer 2

A little hint: $6*6\bmod 10 = 6$.