NO. Such a repunit $11\ldots 1$ is equal to $\frac {10^n - 1} {9}$ for some $n>1$; if it also is some $m^5$ we will have $10^n = 9m^5 + 1$. So $9m^5 + 1 \equiv 0 \pmod{5}$. Since clearly $5 \nmid m$, we have $9m^5 + 1 \equiv -m + 1 \equiv 0 \pmod{5}$ by Fermat's little, so $m\equiv 1 \pmod{5}$. But then $9m^5 + 1 = 9((m-1) + 1)^5 + 1 \equiv 9 + 1 = 10 \not \equiv 0 \pmod{5^2}$, so there are no such repunits.