Is there an elementary proof that
$$\sum_{n=1}^\infty \frac1{n^s} = \sum_{n=1}^\infty \frac1{s^n}$$
Has no real solutions $s>1$?
I tried to prove by induction that for each term in the RHS, we may take arbitrarily many arbitrary terms from the LHS which are greater than the term from the RHS, though it got very messy very quickly. I would consider induction the limit of 'elementary' methods to prove the equation has no solutions.
Motivation: the only solution I could come up with involves some pretty strong theorems. Having to call on them for something that appears so simple leaves me unsatisfied.
Sketch of my proof:
The LHS is the Riemann zeta function $\zeta(s)$. Notice that for $|s|>1$, we have $\sum_{n=1}^\infty \frac1{s^n}=\frac1{s-1}$. So, we are looking for the real $s>1$ such that $\zeta(s)=\frac1{s-1}$. Because $s$ is real, this means we are looking for a $u=s-1$ such that $\int_0^1\frac{\text{Li}_u(t)}{t}dt=\frac1u$ for real $u>0$. But this has no positive real solutions.
No idea of what you mean with $\int_0^1\frac{\text{Li}_u(t)}{t}dt=\frac1u$
$$\zeta(s)-\frac1{s-1} = \sum_{n\ge 1}( n^{-s}-\int_n^{n+1} x^{-s}dx) = \sum_{n\ge 1} \int_n^{n+1}(n^{-s}- x^{-s})dx > 0 $$ Since $$\int_n^{n+1}(n^{-s}- x^{-s})dx=\int_n^{n+1}\int_n^x st^{-s-1}dtdx=O(n^{-s-1})$$ this stays true by analytic continuation for $s > 0$