Calculate the sum of inverse values of ${n\choose 0}, {n\choose 1}, ... {n\choose n}$

A good moment for exploiting the beauty of Euler's Beta function. $$\begin{eqnarray*}\sum_{k=0}^{n}\frac{(-1)^k}{\binom{n}{k}}&=&(n+1)\sum_{k=0}^{n}(-1)^k \frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+2)}\\&=&(n+1)\sum_{k=0}^{n}(-1)^kB(k+1,n-k+1)\\&=&(n+1)\int_{0}^{1}\sum_{k=0}^{n}(-1)^k (1-x)^k x^{n-k}\,dx\\&=&(n+1)\int_{0}^{1}(-1)^n(1-x)^{n+1}+x^{n+1}\,dx\\&=&\color{blue}{\frac{n+1}{n+2}((-1)^n+1)}.\end{eqnarray*}$$

I found the solution, though it is not mine. Write $${n\choose k}' := {(-1)^k\over {n\choose k}} $$

Then it is easy to prove $${n\choose k+1}'-{n\choose k}' =-{n+1\over n}{n-1\choose k}'$$ so we have: \begin{eqnarray} B&=&\sum _{k=0}^n {n\choose k}' \\&=&-{n+1\over n+2} \sum _{k=0}^n \Big[ {n+1\choose k+1}'-{n+1\choose k}'\Big] \\ &=& -{n+1\over n+2}\Big[ {n+1\choose n+1}'-{n+1\choose 0}'\Big] \\ &=& -{n+1\over n+2}\big( (-1)^{n+1}-1\big)\\ &=& {n+1\over n+2}\big( (-1)^n+1\big)\\ \end{eqnarray}