Showing that the last digit of $a$ and $a^{13}$ are the same 
For $a \in \mathbb N$, show that the last digit of $a$ and $a^{13}$
  are the same.

For example: $2^{13} =  8,192$
           $7^{13} = 96,889,010,407$

The last digit of $a^2$ in the set $\{1,2,3,4,5,0\}$ is
$\{1, 4, 9, 6, 5, 0\}$ (due to redundancy, I omit $6\le a\le 9$). Then
for $a^4$ we get $\{1,6,1,6,5,0\}$, the same for $a^8,a^{12}$ and so forth...
Now multiply $a^4$ once more to get 
$$
\{1,6,1,6,5,0\}\cdot\{1,2,3,4,5,10\}=
\{1\cdot 1\; ,6\cdot2\; ,3\cdot1\; ,4\cdot 6\; ,5\cdot 5\; ,0\cdot 10 \}_{\bmod 10}
=\{1,2,3,4,5,0\}
$$
(think of $\cdot$ as Hadamard product) so you already get your last digit back for $a^5$.
