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What constant $c$ will make this equality valid for any $N$ chosen?

$$ \sum_{k=2}^{N}c^{\frac{1}{k\log k}}=N. $$

I tried getting a rough idea of what $c$ should be and got about $1.46$ when $N=1000$ but I don't have much perspicacity on how to proceed.

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I do not think that there is a single value of $c$. So, let us consider that you look for the zero of function $$f(c_n)=\sum_{k=2}^{n}c_n^{\frac{1}{k\log k}}-n$$ which does not make much problems to solve numerically.

For illustration purposes, let $n=10^k$; computing, we should obtain the following values $$\left( \begin{array}{cc} k & c_{10^k} \\ 1 & 1.7140673 \\ 2 & 1.4978998 \\ 3 & 1.4189990 \\ 4 & 1.3760092\\ 5 & 1.3480802 \end{array} \right)$$ I gave up for $k=6$ (my computer too !).

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