Can the cube of an integer end with one 1?

Can it end with 2 ones? three 1's? (Experiment)

We can show by induction that there always exists a number ending in k 1's.

Suppose N < 10^k-1 and N³ ends in at least (k-1) 1's. Then there exists A, 0<=A<10, such that A*10^k-1 + N³ ends with k 1's .

But there exists (prove this!) B with 0<=B<10 such that 3B ends in A. Show that [B*10^k-1 + N]³ ends in k 1's. Hint: binomial expansion.

Bonus question: why does this NOT work for squares? What is the next power it does work for?

Good Hunting!
RD