Are there nontrivial perfect powers of integers that are nontrivial repdigits? For example, $6^5=7776$ is close, but not quite a repdigit. Heuristically, it seems to me that there should not be any, because the longer a number with  (effectively) random digits is, the less the odds of that number being a repdigit become. So as the integer power in question gets bigger, the odds of it being a repdigit fall off quickly.

I can eliminate a few cases.  Repdigits look like $a(111\cdots1)$.  If $a=2$, then you have
$$2(111\cdots1) = b^k.$$
Since the second factor is odd, at most one factor of $2$ appears in $b^k$, a contradiction since $k>1.$
Similarly $a$ can't be $6.$
If $a=4$, then similar reasoning forces $k=2$ and $b=2c.$ Then you have 
$$4(111\cdots1) = (2c)^2$$
$$111\cdots1 = c^2$$
which is impossible the left side is congruent to $3$ modulo $4.$
If $a=8$, it (similarly) forces $k=3$ the equation reduces to 
$$111\cdots1 = c^3.$$
This was proved impossible here:
https://mathoverflow.net/questions/132011/can-repunits-be-perfect-cubes
So $a$ can't be even.  
Now things get fuzzy.  If $a$ is odd and prime, it seems like writing
$$p(111\cdots1) = b^k$$
should get us somewhere.  Certainly it shows that $p^{k-1}$ divides $111\cdots1.$  But I'm stuck there.   That would eliminate $3, 5,$ and $7$.  
That's all I got.
