Bounding the Gamma Function I'm trying to verify a bound for the gamma function
$$ \Gamma(z) = \int_0^\infty e^{-t}t^{z - 1}\;dt. $$
In particular, for real $m \geq 1$, I'd like to show that
$$ \Gamma(m + 1) \leq 2\left(\frac{3m}{5}\right)^m. $$
Knowing that the bound should be attainable, my first instinct is to split the integral as
$$ \Gamma(m + 1) = \int_0^{3m/5} e^{-t}t^{m}\;dt + \int_{3m/5}^\infty e^{-t}t^m\;dt
\leq (1 - e^{-3m/5})\left(\frac{3m}{5}\right)^m + \int_{3m/5}^\infty e^{-t}t^m\;dt. $$
Using integration by parts,
$$ \int_{3m/5}^\infty e^{-t}t^m\;dt = e^{-3m/5}\left(\frac{3m}{5}\right)^m + m\int_{3m/5}^\infty e^{-t}t^{m-1}\;dt.$$
So the problem has been reduced to showing
$$ m\int_{3m/5}^\infty e^{-t}t^{m-1}\;dt \leq \left(\frac{3m}{5}\right)^m. $$
But this doesn't seem to have made the problem any easier.
Any help is appreciated, thanks.

I'll prove something that's close enough for my applications; in particular, that $$\Gamma(m + 1) \leq 3\left(\frac{3m}{5}\right)^m.$$
Let $0 < \alpha < 1$ be chosen later.  We'll split $e^{-t}t^m$ as $(e^{-\alpha t}t^m)e^{-(1 - \alpha)t}$ and use this to bound the integral.
First, take a derivative to find a maximum for $e^{-\alpha t}t^m$.
$$\frac{d}{dt}e^{-\alpha t}t^m = -\alpha e^{-\alpha t}t^m + me^{-\alpha t}t^{m-1} = -\alpha e^{-\alpha t}t^{m - 1}\left(t - \frac{m}{\alpha}\right). $$
So $t = m / \alpha$ is a critical point, and in particular a maximum (increasing before and decreasing after, if you like).
Then we can bound the integral
$$ \Gamma(m + 1) = \int_0^\infty e^{-t}t^m\;dt \leq \left(\frac{m}{\alpha e}\right)^m \int_0^\infty e^{-(1 - \alpha)t}\;dt = \left(\frac{m}{\alpha e}\right)^m \left(\frac{1}{1 - \alpha}\right).$$
Choosing $\alpha = 5/(3e)$ and noting that $\frac{1}{1 - 5/(3e)} \leq 3$, we've proven
$$ \Gamma(m + 1) \leq 3\left(\frac{3m}{5}\right)^m. $$
