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.

If you multiply any two integers whose last digit is 6, you get an integer whose last digit is 6:
$$
\begin{array}
{} & {} & {} & \bullet & \bullet & \bullet & \bullet & \bullet & 6 \\
\times & {} & {} &\bullet & \bullet & \bullet & \bullet & \bullet & 6 \\
\hline
{} & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & \bullet & 6
\end{array}
$$
(Get 36, and carry the "3", etc.)
To put it another way, if the last digit is 6, then the number is $(10\times\text{something}) + 6$.  So
$$
\begin{align}
& \Big((10\times\text{something}) + 6\Big) \times \Big((10\times\text{something}) + 6\Big) \\
= {} & \Big((10\times\text{something})\times (10\times\text{something})\Big) \\
& {} + \Big((10\times\text{something})\times 6\Big) + \Big((10\times\text{something})\times 6\Big) + 36 \\
= {} & \Big(10\times \text{something}\Big) +36 \\
= {} & \Big(10\times \text{something} \Big) + 6.
\end{align}
$$
