You should use Stirling formula as
$k!\approx \sqrt{2\pi k}k^k e^{-k}$
and you will get
$S_k\approx \frac{k^\frac{1}{2}}{\sqrt{2\pi}}t^ke^{-k(t-1)}$
that can be rewritten as
$S_k\approx \frac{1}{\sqrt{2\pi\frac{1}{k}}}e^{-\frac{(t-1)}{\frac{1}{k}}+k\ln t}$
and put $\epsilon=\frac{1}{k}$. Now consider a compact support function $f(t)$ (this means that this function goes fastly enough to zero when its argument $t$ goes to $\pm\infty$). Then
$\lim_{\epsilon\rightarrow 0}\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi\epsilon}}e^{-\frac{t-1}{\epsilon}+\frac{1}{\epsilon}\ln t}f(t)dt.$
Now, applying saddle point method we take the derivative of the argument of the exponential. You will get an extremum for $t=1$. Expanding $-(t-1)+\ln t$ till second order and integrating you will get the proof of your assertion, after extracting $f(1)$.