Proving a combinatoric proof using the result of $\sum_{k=0}^n \binom nk\ (-1)^k = 0$

Question

so I am supposed to solve a proof which seems fairly easy, but the negative exponents in $$\sum_{k=0}^n \binom nk\ (\frac{(-1)^k}{k+1})= \frac{1}{n+1}$$ are making this question very difficult for me. I have tried using binomial theorem on the right side with $(n+1)^{-1}$ but I understand that operation does not make sense. I can also tell that the only difference between the two proofs is that the left side has an additional $(k+1)^{-1}$ and the right side has an additional $(n+1)^{-1}$, but I am still having difficulty solving this question. Hints appreciated.

Answer 1

Hint: Consider antidifferentiating $f(x)=(1-x)^n$.

Answer 2

The binomial theorem gives: $$(1-x)^n= \sum_{k=0}^{n} (-1)^k x^k$$ Integrate w.r.t. x both sides from $x=0$ to $x=1$, to get $$\left .\sum_{k=0}^{n} (-1)^k {n \choose k}\frac{x^{k+1}}{k+1} \right|_{0}^{1}= -\left.\frac{(1-x)^{n+1}}{n+1} \right|_{0}^{1}=\frac{1}{n+1}.$$