Proving that $\int_{0}^{1}x^{n-1}(1-x)^n \mathrm dx =\frac{1}{n {2n\choose n}}$

Let's put $$I_{k,n} = \int_0^1 x^k (1-x)^n dx.$$ By integration by parts we have: $$I_{k,n} = \frac{n}{k+1} I_{k+1, n-1}. $$ Now we can iterate this to get $$ I_{k,n} = \frac{n}{k+1} I_{k+1,n-1} = \frac{n}{k+1} \frac{n-1}{k+2} I_{k+2, n-2} = \ldots = \frac{n!}{(k+1)(k+2) \ldots (k+n)} I_{k+n,0}. (*)$$

We have $$I_{k+n,0} = \int_0^1 x^{k+n} dx = \frac{1}{k+n+1}.$$ Now it remains to write the fraction from $(*)$ as a binomial coefficient.