I want to prove the following:
$$ \sum_{k=1}^n (-1)^{k-1}{n\choose k} \frac{1}{k}=\sum_{k=1}^n\frac{1}{k} $$
I've tested the n=1, ..., 12 cases and they work, which led me to go for induction but I got an extra factor of $\dfrac{N+1}{N+1-k}$ that I had no idea what do to with. I've also tried applying summation by parts, but I cannot seem to choose the correct product of sequences. I know that they are equivalent because both series are solutions to the integral:
$$ -n\int_0^1 x^n\ln(1-x) \;dx $$
I can show the equivalency through the integral above but I am hoping for a more direct proof without going through integration.
