$\lim_{n\to\infty}\frac{n!}{(n-k)!k!}(\frac{\lambda t_{\alpha}}{n})^k(1-\frac{\lambda t_{\alpha}}{n})^{n-k}$ Stiriling formula 
I am supposed to study the behaviour of $\lim_{n\to\infty}\frac{n!}{(n-k)!k!}(\frac{\lambda t_{\alpha}}{n})^k(1-\frac{\lambda t_{\alpha}}{n})^{n-k}$ making use of Stirling formula: $n!\sim \sqrt{2\pi}n^n\sqrt{n}e^{-n}$.

After replacing the $n!$ by the Stirling formula I got a complex expression I cannot simplify.
After commentary here is the expression: $\frac{n^n\sqrt{n}e^{-n}}{(n-k)^{n-k}\sqrt{n-k}e^{-n+k}}(\frac{\lambda t_{\alpha}}{n})^k(1-\frac{\lambda t_{\alpha}}{n})^{n-k}$ 
Question:
1) How should I proceed to compute the above limit using the formula for $n!$?
Thanks in advance!
 A: \begin{align}
& \frac{n^n\sqrt{n}e^{-n}}{(n-k)^{n-k}\sqrt{n-k} \, e^{-n+k} \, k!} \left( \frac{\lambda t_{\alpha}}{n} \right)^k \left( 1-\frac{\lambda t_{\alpha}}{n} \right)^{n-k} \\
&= \frac{(\lambda t_\alpha)^k}{k! \, e^k} \, \frac{\sqrt{n}}{\sqrt{n-k}} \, \frac{(n-\lambda t_\alpha)^{n-k}}{(n-k)^{n-k}} \\
&= \frac{(\lambda t_\alpha)^k}{k! \, e^k} \, \underbrace{\frac{\sqrt{n}}{\sqrt{n-k}} \, \left( \frac{n-\lambda t_\alpha}{n-k}\right)^{-k}}_{\to 1 \text{ as } n \to \infty} \, \left( \frac{n-\lambda t_\alpha}{n-k}\right)^{n} \\
&= \frac{(\lambda t_\alpha)^k}{k! \, e^k} \, \underbrace{\frac{\sqrt{n}}{\sqrt{n-k}} \, \left( \frac{n-\lambda t_\alpha}{n-k}\right)^{-k}}_{\to 1 \text{ as } n \to \infty} \, \left( 1+\frac{1}{(n-k)/(k-\lambda t_\alpha)}\right)^{n} \\
&\xrightarrow[n \to \infty]{} \frac{(\lambda t_\alpha)^k}{k! \, e^k} \, e^{k-\lambda t_\alpha} \\
&= \frac{(\lambda t_\alpha)^k}{k! \, e^{\lambda t_\alpha}}
\end{align}
Remarks: This can be done without using Stirling's Formula.  You may refer to another related question Limit of a probability for details.
