An observation about $x^n$ when $n$ is a positive integral power of $5$ Are there any other $n$ except the positive integral powers of $5$ such that the last digit of $x^n$ is the same as the last digit of $x$ for all positive integers $x$.
 A: In base 10, in order for $x^n$ to have the same final digit as $x$ (what I assume you meant to type), we gather requirements for $n$ by looking at each possible final digit.
0 and 1 place no requirements on $n$. (That is, if $x$ ends in 0 or 1, then so does $x^n$ regardless of $n$)
2 adds the requirement that $n$ be one more than a multiple of $4$. (since if $x$ ends in 2, then $x^n$ ends in 2, 4, 8, 6, 2, ...)
3 adds the same requirement (that $n$ be one more than a multiple of $4$).
4 adds the requirement that $n$ be odd.
5 and 6 place no requirement on $n$.
7 requires that $n$ be one more than a multiple of $4$. So does 8.
9 requires that $n$ be odd.
So in fact, any $n$ that is one more than a multiple of 4 should work, and indeed:
1^13 =             1
2^13 =          8192
3^13 =       1594323
4^13 =      67108864
5^13 =    1220703125
6^13 =   13060694016
7^13 =   96889010407
8^13 =  549755813888
9^13 = 2541865828329

A: Any exponent of the form $n = 4k + 1$ will work: $n = 1, 5, 9, \ldots$.  (In particular all integral powers of 5 are of this form.)
