For the first one, if you write out all the binomial coefficients and cancel identical factors on both sides, you're left with
$\frac1{(i+k)!}\le\frac1{i!}\frac1{k!}\;,$
which is clearly true.
For the second one, you can use Stirling's approximation in the form
$\sqrt{2\pi}\ n^{n+1/2}\mathrm e^{-n} \le n! \le \mathrm e\ n^{n+1/2}\mathrm e^{-n}\;,$
which for $k(n-k)\gt0$ yields
$ \begin{align} \binom nk &=\frac{n!}{k!(n-k)!} \\ &\le \frac{\mathrm e\ n^{n+1/2}\mathrm e^{-n}}{\sqrt{2\pi}\ k^{k+1/2}\mathrm e^{-k}\sqrt{2\pi}\ (n-k)^{n-k+1/2}\mathrm e^{-(n-k)}} \\ &= \frac{\mathrm e}{2\pi}\sqrt{\frac{n}{k(n-k)}}\frac{n^n}{k^k(n-k)^{n-k}} \\ &\le \frac{n^n}{k^k(n-k)^{n-k}}\;. \end{align} $
For $k(n-k)=0$, equality holds in your inequality if we interpret $0^0$ as $1$.