TITLE: Prove: $\binom{n}{k}^{-1}=(n+1)\int_{0}^{1}x^{k}(1-x)^{n-k}dx$ for $0 \leq k \leq n$
QUESTION [36 upvotes]: I would like your help with proving that for every  $0 \leq k \leq n$, 

$$\binom{n}{k}^{-1}=(n+1)\int_{0}^{1}x^{k}(1-x)^{n-k}dx
 . $$

I tried to integration by parts and to get a pattern or to use the binomial formula somehow, but it didn't go well.
Thanks a lot!

REPLY [6 votes]: The equation is equivalent to 
$$\binom{n+m}{n}^{-1}=(n+m+1)\int_{0}^{1}x^{n}(1-x)^{m}dx$$
with 
$$ m\ge0 , n\ge 0$$
Or
$${1\over(n+m+1)!}=\int_{0}^{1}{x^{n}(1-x)^{m}\over n!m!}dx$$
And hence one can multiply the two series and integrate from 0 to 1:
$$e^{tx} = \sum_{n=0}^\infty{t^nx^n\over n!}$$
$$e^{s(1-x)} = \sum_{m=0}^\infty{s^m(1-x)^m\over m!}$$
Giving
$$\int_{0}^{1}{e^{(t-s)x+s}}dx=\sum_{m,n=0}^\infty{t^ns^m\int_{0}^{1}{x^{n}(1-x)^{m}\over n!m!}dx}$$
Finally, the LHS is equal to $$e^t-e^s\over(t-s)$$
And using the exponential series and the fact that 
${{t^n-s^n}\over{t-s}} = t^{n-1}+t^{n-2}s+...+ts^{n-2}+s^{n-1}$
gives the result.