How can I prove that: $\forall n\geq 2 $ $ \sum_\limits{k=2}^{n}\frac{1}{\log_{k}{n!}}=1$

Question

$\forall n\geq 2 $ $ \sum_\limits{k=2}^{n}\frac{1}{\log_{k}{n!}}=1$

I worked it out for $n=3$:

$\sum_\limits{k=2}^{n}\frac{1}{\log_{k}(n!)}=\frac{1}{\log_{2}(3)+\log_{2}(2)}+\frac{1}{\log_{3}(3)+\log_{3}(2)}=\frac{1}{\log_{2}(3)+1}+\frac{1}{1+\log_{3}(2)}$

$=\frac{1}{\log_{2}(3)+1}+\frac{1}{1+\frac{1}{\log_{2}(3)}}=\frac{1}{1+\frac{1}{\log_{2}(3)}}\left ( \frac{1}{\log_{2}(3)}+1 \right )=1$

But I fail to see how to generalize the result. Thank you

Answer 1

HINT:

$$\log_k(n!)=\frac{\log(n!)}{\log(k)} \tag 1$$

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SPOILER ALERT: Scroll over the highlighted area to reveal the solution

Using $(1)$, we can write $$\begin{align}\sum_{k=1}^n\frac{1}{\log_k(n!)}&=\sum_{k=1}^n\frac{\log(k)}{\log(n!)}\\\\&=\frac{\sum_{k=1}^n\log(k)}{\log(n!)}\\\\&=\frac{\log(n!)}{\log(n!)}\\\\&=1\end{align}$$as was to be shown!