Show that $\Gamma$ is well defined and that for all $y > 0 , y \Gamma ( y ) = \Gamma ( y + 1 )$. Deduce that, for all $n \in \mathbb { N } , \Gamma ( n + 1 ) = n !$.
Recall that $\Gamma : ] 0 , + \infty [ \rightarrow \mathbb { R }$ is defined by $\Gamma ( y ) = \int _ { 0 } ^ { \infty } e ^ { - t } t ^ { y - 1 } d t$ and that $\Gamma \left( \frac { 1 } { 2 } \right) = \sqrt { \pi }$.

Show that for all $y > 0$, we have $\Gamma ( y ) = y ^ { - 1 } \int _ { 0 } ^ { + \infty } e ^ { - t } t ^ { y } d t$, then that
$$\Gamma ( y ) = e ^ { - y } y ^ { y } \int _ { - 1 } ^ { + \infty } e ^ { - y \phi ( s ) } d s$$
where $\phi$ is the function defined on $] - 1 , + \infty [$ by $\phi ( s ) = s - \ln ( 1 + s )$.
Recall that $\Gamma : ] 0 , + \infty [ \rightarrow \mathbb { R }$ is defined by $\Gamma ( y ) = \int _ { 0 } ^ { \infty } e ^ { - t } t ^ { y - 1 } d t$.