An elegant proof

I'm taking this proof from Introductory Combinatorics, Third Edition by Kenneth P. Bogart. In the first week of classes I have discovered an elegant proof that I would very much like to share with you all.

The powers of two are all even and the powers of five all end in five. Show that for any other prime p, among the first six powers of p, one of these numbers must end in one.

If a prime is not two, then no power of it is even, so a power can only end in one of the five odd digits. Then the function given by

f(i) = the last digit of pⁱ

from {1, 2, 3, 4, 5, 6} to the set of digits {1, 3, 5, 7, 9} cannot be one-to-one. Therefore pⁱ and p^j have the same last digit for some i and j between 1 and 6 with i < j. Thus p^{j -} pⁱ ends in a zero, so it is a multiple of 10. Thus pⁱ⁽p^j-i - 1) is a multiple of 10, and since neither 2 nor 5 is a factor of p, (p^j-i - 1) must be a multiple of 10 (and therefore end in zero), so adding 1 to each side tells us that p^j-i ends in a 1.