Gamma Function is always positive

Question

To prove that the $\Gamma (s) > 0$ for all $s > 0$ we consider the following, $\Gamma(s) \geq \int\limits_0^1 e^{-t} t^{s-1} dt \geq e^{-1} \int\limits_0^1 t^{s-1} dt$. I do not know how to show the second inequality.

Answer 1

If $f \geq g$ in $[0,1]$, then $\int\limits_0^1 f(t) dt \geq \int\limits_0^1 g(t) dt$. In the second inequality $f(t) = e^{-t}t^{s-1}$, and $g(t) = e^{-1} t^{s-1}$. The latter is smaller, because the function $e^{-t}$ decreases in $[0,1]$ and its smallest value there is $e^{-1}$.