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Are there any $k\gt2$ for which we have a solution of the nonlinear diophantine equation $x^{k-1}=\sum_{i=0}^{k-1}10^i$.

This question arose when I tried to provide a simple solution to this question Find matrix $A\in \mathcal{M}_n (\mathbb{N})$ such that $A^k =\left( \sum_{i=1}^{k}10^{i-1} \right)A$. for at least some $k\in[2,\infty)\cap\Bbb N$.

I tried the first nine or ten, and there were none.

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    Apart from the trivial solution when $k=2$, I think you can show that $10<x<11$, so there are no integer solutions. – Jaap Scherphuis Sep 30 '20 at 07:45
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    I am not sure whether it has been proven, but the rep-units $R_n=\frac{10^n-1}{9}$ for $n>1$ should be not even perfect powers. – Peter Sep 30 '20 at 08:11

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Taking @Jaap Scherphuis' suggestion, note that $10^{k-1}\lt\underbrace{111\dots1}_{\text{k-times}}\lt11^{k-1}$.

The first inequality is obvious. For the second, just apply the binomial theorem to $(10+1)^{k-1}$.

Thus there are no solutions.