sequences and series - How to prove $sumlimits_{k=0}^{N} frac{(-1)^k {N choose k}}{(k+1)^2} = frac{1}{N+1} sumlimits_{n=1}^{N+1} frac{1}{n}$

In the process of proving a more complicated relation, I've come across the following equality that I'm having trouble proving:

$$\sum\limits_{k=0}^{N} \frac{(-1)^k {N \choose k}}{(k+1)^2} = \frac{1}{N+1} \sum\limits_{n=1}^{N+1} \frac{1}{n}$$

I was already able to prove the following similar equality:

$$\sum\limits_{k=0}^N \frac{(-1)^k {N \choose k}}{k+1} = \frac{1}{N + 1}$$

but I'm unsure how to proceed with the first one. I assume it has something to do with the fact that every term in the left hand side of the first equality is $\frac{1}{k+1}$ times a term in the left hand side of the second equality. Any help would be greatly appreciated.