Prove that $\Gamma(p) \cdot \Gamma(1-p)=\frac{\pi}{\sin (p\pi)}$ for $p \in (0,\: 1)$ Prove that $$\Gamma(p)\cdot \Gamma(1-p)=\frac{\pi}{\sin (p\pi)},\: \forall  p \in (0,\: 1),$$
where $$\Gamma (p)=\int_{0}^{\infty} x^{p-1} e^{-x}dx.$$
Here's what I tried:
We have
$$B(p, q)=\int_{0}^{1} x^{p-1} (1-x)^{q-1}dx=\frac{\Gamma(p)\cdot \Gamma(q)}{\Gamma(p+q)}$$
Hence
$$B(p, 1-p)=\frac{\Gamma(p)\cdot \Gamma(1-p)}{\Gamma(1)}=\Gamma(p)\cdot \Gamma(1-p)=\int_{0}^{1} x^{p-1} (1-x)^{-p}dx$$
But from here I don't know how to proceed.

Another definiton of Gamma function is
$$\Gamma(p)=\lim_{n\rightarrow \infty}\dfrac{n!\ n^p}{p(p+1)\cdots(p+n)}$$
Then, $\Gamma(x)\Gamma(1-x)$ is:
$$\Gamma(x)\Gamma(1-x)=\lim_{n\rightarrow \infty}\dfrac{n!\ n^x}{x(x+1)\cdots(x+n)}\dfrac{n!\ n^{1-x}}{(1-x)(2-x)\cdots(1-x+n)}$$
Simplifying and reorganizing:
$$\Gamma(x)\Gamma(1-x)=\lim_{n\rightarrow \infty}\dfrac{(n!)^2 \ n}{x(x+1)(1-x)(x+2)(2-x)\cdots(n-x)(n+x)(1-x+n)}$$
$$=\lim_{n\rightarrow \infty}\dfrac{(n!)^2 \ n}{x(n-x+1)\prod\limits_{i=1}^n (i^2-x^2)}=\lim_{n\rightarrow \infty}\dfrac{(n!)^2 \ n}{x(n-x+1)\prod\limits_{i=1}^n i^2(1-\dfrac{x^2}{i^2})}$$
$$= \lim_{n\rightarrow \infty}\dfrac{(n!)^2 \ n}{x(n-x+1)(n!)^2\prod\limits_{i=1}^n (1-\dfrac{x^2}{i^2})}=\Bigg(\lim_{n\rightarrow \infty}\dfrac{n}{n-x+1}\Bigg)\Bigg(\lim_{n\rightarrow \infty}\dfrac{1}{x\prod\limits_{i=1}^n (1-\dfrac{x^2}{i^2})} \Bigg)$$
The first limit is one, so,
$$\Gamma(x)\Gamma(1-x)=\lim_{n\rightarrow \infty}\dfrac{1}{x\prod\limits_{i=1}^n (1-\dfrac{x^2}{i^2})}$$
But, remember that we can write $\sin(\pi x)$ as
$$\sin(\pi x)=\pi x \prod\limits_{i=1}^\infty(1-\dfrac{x^2}{i^2})$$
Then,
$$\Gamma(x)\Gamma(1-x)=\dfrac{\pi}{\pi x\prod\limits_{i=1}^\infty (1-\dfrac{x^2}{i^2})}=\dfrac{\pi}{\sin(\pi x)}$$
What we wanted to show.
