Proof of $\sum _{k=1}^{n-1}k\ \log \left(\frac{k+1}{k}\right)\ =\ n \log \left(n\right) - \log \left(n!\right)$ via induction or directly How can one prove
$$\sum _{k=1}^{n-1}k\ \log \left(\frac{k+1}{k}\right)\ =\ n \log \left(n\right) - \log \left(n!\right)$$
for $n \in \mathbb{N}$ with $n \geq 2$?
I know that by telescopic sum we get
$$k \sum\limits_{k=1}^{n-1} \log(\frac{k+1}{k}) =k \sum\limits_{k=1}^{n-1}\log(k+1)-\log(k) =\log (n)$$
and that $$\log(\frac k {k+1}) = \log(k) - \log(k+1) = -(\log(k+1) - \log(k)) = - \log(\frac {k+1} k)$$
but that's it.
Can one prove the equality above directly or via induction?

$$\begin{align}
\sum_{k=1}^{n-1} k\ln\left(\frac{k+1}{k} \right)&=\sum_{k=1}^{n-1} k\ln(k+1)-\sum_{k=1}^{n-1}k\ln(k)\\
\\
&=\sum_{k=1}^{n-1} (k+1-1)\ln(k+1)-\sum_{k=1}^{n-1}k\ln(k)\\
\\
&=\sum_{k=1}^{n-1} (k+1)\ln(k+1)-\sum_{k=1}^{n-1}\ln(k+1)-\sum_{k=1}^{n-1} k\ln(k)\\
\\
&=n\ln(n)-\sum_{k=1}^{n-1}\ln(k+1)\\
\\
&=n\ln(n)-\ln(n!)
\end{align}$$
