We can't have a relationship of the form $$\forall n\in\mathbb N^*, \sum_{k=1}^nk^a=\left(\sum_{k=1}^nk^b\right)^c$$ for $a,b,c\in\mathbb N$, except in the case $c=1$ and $a=b$ or $a=3$, $b=1$ and $c=2$. Indeed, we can write $$\sum_{k=1}^nk^a =n^{a+1}\frac 1n\sum_{k=1}^n\left(\dfrac kn\right)^a$$ hence $$\sum_{k=1}^nk^a\;\overset{\scriptsize +\infty}{\large \sim}\;n^{a+1}\int_0^1t^adt=\dfrac{n^{a+1}}{a+1}$$ and if we have the initial equality we should have $a+1 =(b+1)c$ and $a+1=(b+1)^c$. In particular, $(b+1)^{c-1}=c$. If $c>1$, then $c= (b+1)^{c-1}\geq 2^{c-1}\geq c$, and we should have $c=2$ and $b=1$, therefore $a=3$.