Prove $\int_{0}^{\infty}e^{-nx}\sin^{2k}xdx={(2k)!\over n\prod_{j=1}^{k}(n^2+4j^2)}$ 
$$I=\int_{0}^{\infty}e^{-nx}\sin^{2k}xdx={(2k)!\over n\prod_{j=1}^{k}(n^2+4j^2)}\tag1$$

Recall
$$\sin^{2k}(x)={1\over 2^{2k}}{2k\choose k}+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\cos[(2k-2j)x]$$
$$I={1\over 2^{2k}}{2k\choose k}\int_{0}^{\infty}e^{-nx}dx+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\int_{0}^{\infty}e^{-nx}\cos[(2k-2j)x]dx\tag2$$
$$I={1\over 2^{2k}n}{2k\choose k}+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\int_{0}^{\infty}e^{-nx}\cos[(2k-2j)x]dx\tag3$$
$$\int_{0}^{\infty}e^{-nx}\cos[(2k-2j)]dx={n\over n^2+(2k-2j)^2}$$
$$I={1\over 2^{2k}n}{2k\choose k}+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\cdot{n\over n^2+(2k-2j)^2}\tag4$$
How can I simplify the LHS sum to the RHS product?
$${1\over 2^{2k}n}{2k\choose k}+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\cdot{n\over n^2+(2k-2j)^2}={(2k)!\over n\prod_{j=1}^{k}(n^2+4j^2)}\tag5$$
 A: Olivier Oloa already indicated two approaches in the comments:
one by repeated integration by parts (worked out in detail in
the answer by B. Mehta), one by checking the partial-fraction
decomposition.  Here is another, via the
Beta function.
Consider more generally
$$
I(a) := \int_0^\infty e^{-nx} \left(\frac{e^{ax}-e^{-ax}}{2a}\right)^{2k} dx
$$
for complex $a$ for which the integral converges, that is, with
$|{\rm Re}(2ka)| < n$.  Then Mr. Bean's integral $I$ is our $I(i)$.
Expanding $(e^{ax}-e^{ax})^k$ and integrating termwise we see that
$I(a)$ is a linear combination of $1/(n+ja)$ over $j=k,k-2,k-4,\ldots,-k$;
in particular, it is a rational function of $a$ and $n$.  Hence it is enough
to evaluate $I(a)$ for real $a$.  But then the change of variables
$u = e^{-2ax}$ gives
$$
I(a) = \frac1{(2a)^{2k+1}} \int_0^1 u^{n/2a} (u^{-1/2} - u^{1/2})^{2k}
\, \frac{du}{u},
$$
which is
$$
\frac1{(2a)^{2k+1}} \int_0^1 u^{\frac{n}{2a}-k-1}_{\phantom0} (1-u)^{2k} \, du
= \frac1{(2a)^{2k+1}} B(2k+1, \frac{n}{2a} - k).
$$
This, in turn, is
$$
\frac{(2k)!}{(2a)^{2k+1}}
  \frac{\Gamma(\frac{n}{2a}-k)}{\Gamma(\frac{n}{2a}+k+1)}
= \frac{(2k)!}{(2a)^{2k+1} \prod_{j=-k}^k (\frac{n}{2a}+j)}.
= \frac{(2k)!}{\prod_{j=-k}^k (n+2aj)}.
$$
Grouping together the $j$ and $-j$ terms, we deduce that
$$
I(a) = \frac{(2k)!}{n\prod_{j=1}^k (n^2-(2aj)^2)}.
$$
The desired result follows on taking $a=i$.  QED
This integral appears as 3.895 #1 on page 478 of
Gradshteyn and Ryzhik
(with $n,k$ called $\beta, m$), attributed to
FI [= Fikhtengolts, Course in differential and integral calculus (1947-1949]
Vol. II, 615, and WA [= Watson, A Treatise on the Theory of Bessel Functions,
2nd ed. (1944)] 620a.  The analogous formula with odd exponent
$$
\int_0^\infty  e^{-nx} \sin^{2k+1} x \, dx
 = \frac{(2k+1)!}{\prod_{j=0}^k (n^2+(2j+1)^2)}
$$
appears as 3.895 #4 on the same page, with the exact same attributions;
it can be obtained in the same way, or using either of
Olivier Oloa's techniques.  Even non-integral exponents
can be accommodated, though one must be careful to choose
the right branches of powers of $\sin x$.
A: Following Olivier's idea in the comments, let $I_k = \int_0^\infty e^{-nx} \sin^{2k}\! x \ dx$ for $k\geq1$.
$$\begin{align}
I_k &= \int_0^\infty e^{-nx} \sin^{2k}\! x \ dx \\
&=\left[-\frac{1}{n}e^{-n x}\sin^{2k}\! x\right]_0^\infty-\int_0^\infty -\frac{2k}{n}e^{-nx} \sin^{2k-1}\! x \cos x\ dx \\
&=\frac{2k}{n} \int_0^\infty e^{-n x}\sin^{2k-1}\!x\cos x\ dx \\
&=\frac{2k}{n}\left(\left[-\frac{1}{n}e^{-n x}\sin^{2k-1}\!x\cos x\right]_0^\infty-\int_0^\infty -\frac{1}{n}e^{-n x}\frac{d}{dx}(\sin^{2k-1}\!x\cos x)\ dx \right) \\
&=\frac{2k}{n^2}\int_0^\infty e^{-n x}\left((2k-1)\sin^{2k-2}\!x \cos^2 x-\sin^{2k}\!x\right)\ dx \\
&=\frac{2k}{n^2}\int_0^\infty e^{-n x}\left((2k-1)\sin^{2k-2}\!x (1-\sin^2 x)-\sin^{2k}\!x\right)\ dx \\
&=\frac{2k}{n^2}\int_0^\infty e^{-n x}\left((2k-1)\sin^{2k-2}\!x-2k\sin^{2k}\!x\right)\ dx \\
&=\frac{2k\times(2k-1)}{n^2}I_{k-1}-\frac{2k \times2k}{n^2}I_{k}\\
n^2 I_k &= 2k(2k-1)I_{k-1}-4k^2I_k \\
(n^2+4k^2)I_k&=2k(2k-1)I_{k-1} \\
I_k &= \frac{2k(2k-1)}{n^2+4k^2} I_{k-1}
\end{align}$$
We also need to check $I_0$:
$$\begin{align}I_0 &= \int_0^\infty e^{-nx}dx \\
&= \left[-\frac{1}{n}e^{-nx}\right]_0^\infty \\
&= \frac{1}{n}\end{align}$$
So, $I_1 = \frac{2\times1}{n^2+4} \frac{1}{n} = \frac{2!}{n^2+4} \frac{1}{n}$, similarly $I_2 = \frac{4\times3}{n^2+4\times2^2} \frac{2!}{n^2+4} \frac{1}{n} = \frac{4!}{(n^2+4\times2^2)(n^2+4)} \frac{1}{n}$, and so on.  You can follow the pattern (or, more formally, use induction) to see that 
$$I_k = \frac{(2k)!}{n \prod_{j=1}^k (n^2 + 4j^2)}$$
