Upper bound of the integral $\frac{a^{n}}{n!}\int_{0}^{1} e^{at} t^{n}(1-t)^{n}dt$. I came across this integral in a proof related to using the Legendre polynomial to establish irrationality.

$$
I_{n}=(-1)^{n} \frac{a^{n}}{n!} \int\limits_{0}^{1} e^{at} t^{n}(1-t)^{n}dt $$

The author then assumed that this upper bound was trivial.
$$|I_{n}|<|\frac{a}{n}|^{n} \frac{e^{a}}{n!} $$
But,I do not understand how this bound was obtained.Perhaps I missed something very simple.I'd be grateful if someone could show how this bound was obtained.
 A: Well, when $0\leq t\leq1$, then $e^{at}\leq e^{a}$, and $t(1-t)=t-t^2$ takes it's maximum at $t=\dfrac12$, it is easy to find the uper bound which is trivial now!
A: Actually, the upper bound was explained earlier by user 108128, but in this case it is much easier to get the closed form answer. 
 Applying the identity (see Prudnikov A. P., Brychkov Yu. A., Marichev O. I. Integrals and Series. Volume 1: Elementary Functions. Gordon & Breach Science Publishers / CRC Press. New York-London.(1986)):
$$\int\limits_{0}^{a} e^{-px} x^{\alpha-1}(a-x)^{\alpha-1}dx=\sqrt{\pi}\Gamma(\alpha)\left(\frac{a}{p}\right)^{\alpha-\frac{1}{2}}e^{\frac{-ap}{2}}I_{\alpha-\frac{1}{2}}\left(\frac{ap}{2}\right)$$
ans after all the simplifications one can get:
$$(-1)^{n} \frac{a^{n}}{n!} \int\limits_{0}^{1} e^{at} t^{n}(1-t)^{n}dt=(-1)^n \sqrt{\frac{\pi }{a}} e^{a/2}I_{n+\frac{1}{2}}\left(\frac{a}{2}\right),$$
where $I_{n+\frac{1}{2}}\left(\frac{a}{2}\right)$-is the modified Bessel function.
Hence for further bounding one can use the Bessel function approximation.
A: $$I_{n}=(-1)^{n} \frac{a^{n}}{n!} \int\limits_{0}^{1} e^{at} t^{n}(1-t)^{n}dt$$
if we consider the part of the function $e^{at}t^n(1-t)^n$ we can see that the maximum of this function in the range $(0,1)$ is always much less than $e^a$ so the integral must also be less that $e^a$ by considering the area under the curve as a rectangle of height $e^a$ and length 1 stated by the limits of the integral.
We could now consider the integral simply as $$I_n<(-1)^n\frac{a^n}{n!}e^a\,\therefore\,|I_n|<\frac{a^n}{n!}e^a$$ then notice that the integral is always less than $\frac{e^a}{n!}$ and approximate $n!=n^n$
This reasoning is close I guess.
