Q:

The final digit of fourth powers

I am working on "Elementary Number Theory" By Underwood Dudley and this is problem 13 in Section 4.
The question is "What can the last digit of a fourth power be?" I got the correct answer but I'm wondering if there is another more elegant way to do it. I'm also wondering if my argument is solid or if I just got the right answer by chance. My argument is as follows:
Any number A can be written as:
$A$ = $10^kn_k$ + $10^{k-1}$$n_{k-1}$ + $...$ + $10n_1$ + $n_0$,
where $n_k$ is the starting digit of the number and can take on any integer from 0 to 9 and so forth.
If we raise this to the fourth power and don't combine any of the terms, each term should be divisible by 10 except maybe not $n_0^4$. Now, $n_0^4$ must end must end in k where k satisfies:
$n_0^4$ $\equiv k\pmod {10}$. 
And since $n_0^4$ ends in k, $A$ must end in k (since all of the other terms in the sum are divisible by 10). So, it suffices to just check what the last digit of the fourth powers of 1 through 9 are (since $0 \leq n_0 \leq 9$):
$0^4$ $\equiv 0\pmod{10}$
$1^4 \equiv 1\pmod{10}$
$2^4 \equiv 6\pmod{10}$
$3^4 \equiv 1\pmod{10}$
$4^4 \equiv 6\pmod{10}$
$5^4 \equiv 5\pmod{10}$
$6^4 \equiv 6\pmod{10}$
$7^4 \equiv 1\pmod{10}$
$8^4 \equiv 6\pmod{10}$
$9^4 \equiv 1\pmod{10}$.
So the fourth power of any integer must end in either a 0,1,5, or 6.
Did I get lucky or is this ok? Are there other more elegant ways?
Thank you!

A:

That looks fine, though you could have shortened the argument by noting that the last digit of a product depends only on the last digit of each factor, and hence the same goes for powers.
You could have shortened it further by considering fourth powers modulo 2 and modulo 5. You can get either 0 or 1 modulo 2 (obviously), and the same goes modulo 5 (slightly less obviously, but you could use Fermat' little theorem if you felt super lazy).
Combining the two results gives you 0, 1, 5, 6 modulo 10 as you have found.

