Let's see nature of $\frac{e^{n}}{n*2^{n}}$.
$\implies=\frac{e^{n}}{n*2^{n}} = e^{ln(e^{n})-ln(n*2^{n})}$
We find that $$lim_{x\to\infty} ln(e^{n})-ln(n*2^{n}) = lim_{x\to\infty}0.7n-lnn = lim_{x\to\infty}n*lim_{x\to\infty}(0.7-\frac{lnn}{n}) $$ is infinite as $n$ approaches to infinite.
$\implies\frac{e^{n}}{n*2^{n}}= \infty $
$\implies e^{n}=\omega(n*2^{n})$