I found a formula $$\left|J_{n}(x)\right| \leq \left(\frac{ex}{2n}\right)^{n},\quad n\in\mathbb{N}_+,\ x>0$$ in a paper without explanation. Here $J_n(x)$ is the Bessel function of the first kind.
I am wondering whether this is true because I found no similar upper bound of Bessel functions on the Internet. I can just associate it with the power series of $J_n$ $$J_{n}(x)=\sum_{m=0}^{\infty} \frac{(-1)^{m}}{m ! (m+n)!}\left(\frac{x}{2}\right)^{2 m+n}$$ and the first term of it is just $$\frac{1}{n!}\left(\frac{x}{2}\right)^{n}< \frac{1}{\sqrt{2\pi n}}\left(\frac{ex}{2n}\right)^{n}$$ which is similar to the right hand side of the very beginning formula. However, the remaining terms of the power series may not be very small that can be discarded.
So does the formula at the very beginning right? If it is right, how to prove it? Any tips or references are welcome!
By http://dlmf.nist.gov/10.14.E4 and the known bound $(n/e)^n < n!$, we have $$ J_n (x) \le \frac{1}{{n!}}\left( {\frac{x}{2}} \right)^n \le \left( {\frac{e}{n}} \right)^n \left( {\frac{x}{2}} \right)^n = \left( {\frac{{ex}}{{2n}}} \right)^n $$ for $n\in \mathbb{N}$ and $x>0$. For the first inequality, see page 49 in G. N. Watson's book A Treatise on the Theory of Bessel Functions.