Copying perfect 8's

$8,88,888,8888,88888, 888888, \ldots$

By testing out the first few numbers above, it appears that the very first term (8) is the only term that is also a perfect cube. But is this actually true?

That is, is it true that $\underbrace{888888\ldots8}_{\text{All its digits are 8's}} = N^3$ has no solution for integer $N> 2$ ?