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comment Why is the last digit of $n^5$ equal to the last digit of $n$?
What I love about this is that not only will $n^1,5,9,...$ end in $n$, but if you observe the in between powers, there is always a four number cycle. e.g. for $3^k, k->\infty$, we observe the numbers $3, 9, 27, 81, 243, 729, 2187, 6561, 19683, ...$, rendering the cycle $3, 9, 7, 1$.