<s>[INST] how to show $B(n,m)=\frac{(n-1)!(m-1)!}{(n+m-1)!}$? 
I have done the following transformation:
$(1-x)^{m-1}=u$ $\Rightarrow -(m-1)(1-x)^{m-2}dx=du$
$x^{n-1}dx=dv$ $\Rightarrow \frac{x^n}{n}=v$
Hence $$\int^1_0x^{n-1}(1-x)^{m-1}dx= (1-x)^{m-1}\frac{x^n}{n}\bigg |^1_0+ \int\frac{x^n}{n}(m-1)(1-x)^{m-2}dx=(1-1)^{m-1}\frac{1}{n}-1^{m-1}\frac{0}{n} + \frac{m-1}{n}\int^1_0x^n(1-x)^{m-2}dx= \frac{m-1}{n}\int^1_0x^n(1-x)^{m-2}dx=\frac{m-1}{n}B(n+1,m-1)$$

But i'm not exactly sure how to show $B(n,m)=\frac{(n-1)!(m-1)!}{(n+m-1)!}$?

 [/INST] Beginning with the result you have shown, repeated application yields the following chain of equalities:
\begin{align*}
B(n,m) &= \frac{m-1}{n}\cdot B(n+1,m-1) \\
&= \frac{m-1}{n}\cdot \frac{m-2}{n+1}\cdot B(n+2,m-2) \\
&= \cdots \\
&= \frac{m-1}{n}\cdot \frac{m-2}{n+1} \cdot \cdots \cdot \frac{1}{n+m-2} \cdot B(n+m-1,1) \\
&= \frac{(m-1)!(n-1)!}{(n+m-2)!}\cdot B(n+m-1,1)
\end{align*}
It remains to calculate $B(n+m-1,1)$: $$B(n+m-1,1) = \int_0^1x^{n+m-2} \, \mathrm{d}x = \left[ \frac{x^{n+m-1}}{n+m-1} \right]_0^1 = \frac1{n+m-1}.$$
Hence, the result follows, namely: $$B(n,m) = \frac{(m-1)!(n-1)!}{(n+m-1)!}$$
</s>