Natural log summation representation

In working through a problem, I've encountered the need to express

$\log n = \sum \limits_{k = 1}^{n - 1} \log(1 + \frac{1}{k})$

where $\log k$ is the natural logarithm of k. I'm fairly certain the above equality holds (please let me know if otherwise), but I'm not quite sure how to derive it. Any help would be greatly appreciated.

So far, I've tried expressing $\log n = \log(n!) - \log((n-1)!)$ to no avail.

Many thanks!

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You were really close to the answer you accepted; if you expanded $\log(n!)$ and $\log((n-1)!)$ separately and resummed them, you'd basically get the given answer in reverse. – qgp07 Mar 31 '14 at 14:35

\begin{align*} \sum_{k=1}^{n-1}\log\left(1+\frac1k\right)&=\sum_{k=1}^{n-1}\log\frac{k+1}k\\\\ &=\sum_{k=1}^{n-1}\big(\log(k+1)-\log k\big)\\ &=\log n-\log 1\\ &=\log n\;. \end{align*}