<s>[INST] 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: $2^{13} =  8,192$
           $7^{13} = 96,889,010,407$
 [/INST] Claim: For all integers $a$ we have $10|a^5-a$.
This follows from Fermat Little Theorem, but can also be proven elementary.
Proof of the claim
$$a^5-a=a(a-1)(a+1)(a^2+1) \,.$$
Now, one of $a$ or $a+1$ is even thus $a^5-a$ is divisible by $2$.
Also
$$a^5-a=a(a-1)(a+1)(a^2-4+5)=a(a-1)(a+1)(a^2-4)+5a(a-1)(a+1)$$
$$a^5-a=(a-2)(a-1)a(a+1)(a+2)+5a(a-1)(a+1)$$
The product of 5 consecutive integers is always divisible by 5, hence so is $a^5-a$.
This proves the claim.
Proof of your exercise
$$a^{13}-a=a^{13}-a^9+a^9-a^5+a^5-a=(a^5-a)(a^8+a^4+1) \,.$$
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