Proof that $6^n$ always has a last digit of $6$

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 $$ \begin{align} & \Big((10\times\text{something}) + 6\Big) \times \Big((10\times\text{something}) + 6\Big) \\ = {} & \Big((10\times\text{something})\times (10\times\text{something})\Big) \\ & {} + \Big((10\times\text{something})\times 6\Big) + \Big((10\times\text{something})\times 6\Big) + 36 \\ = {} & \Big(10\times \text{something}\Big) +36 \\ = {} & \Big(10\times \text{something} \Big) + 6. \end{align} $$

We can prove it using mathematical induction.

Claim: $6^n\equiv 6\bmod 10$ for all $n\in\mathbb{N}$ (the symbol $\mathbb{N}$ denotes the natural numbers, and $\bmod 10$ means we are using modular arithmetic with a modulus of 10).

Base case (i.e., showing it's true for $n=1$): $$6^1\equiv 6\bmod 10\qquad\checkmark$$

Induction step (i.e., showing that, if it is true for $n=k$, then it is true for $n=k+1$):

$$6^k\equiv 6\bmod 10\implies 6^{k+1}\equiv 6^k\cdot 6\equiv6\cdot 6\equiv 36\equiv 6\bmod 10\qquad\qquad\checkmark$$

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 $\rm\:(mod\ m)\:.\:$