Limit of $\frac{n!}{k! \times n^{k+1}}$ as $n$ approaches infinity What will be the solution to the following limit with $k$ as any constant? 
$$\lim_{n \to \infty} \frac{n!}{k! \times n^{k+1}}$$
This is what I've tried so far:
Since $k$ is a constant, we can rewrite it like:
$$\frac{1}{k!}\lim_{n \to \infty} \frac{n!}{n^{k+1}}$$
Like seriously, I can't think of what do next!
I'd really like some help. Thanks! 
 A: Hint: 
For $n > k +1,$
$$\frac{n!}{n^{k+1}} = (n-k-1)!\prod_{j=1}^k(1 - j/n) > (n- k-1)\prod_{j=1}^k(1 - j/n), \\ \prod_{j=1}^k(1 - j/n) \to 1 \implies \prod_{j=1}^k(1 - j/n) > 1/2 \,\,\text{for sufficiently large} \,\,n$$
A: Method 1...Let $n'=n/2$ if $n$ is even and $n'=(n-1)/1$ if $n$ is odd. Then $n'\geq (n-1)/2$ and $n-n'\geq n/2$ so we have  $$n!\geq (n')!\cdot (n')^{n-n'}\geq (n')^{n-n'}\geq \left(\frac {n-1}{2}\right)^{n/2}.$$ Therefore for $n\geq 2$  we have $$\frac {n!}{k!n^{k+1}}\geq \frac {(\frac {n-1}{2})^{n/2}}{k!n^{k+1}}=$$ $$=\left(\frac {n-1}{2}\right)^{(n/2-k-1)}\cdot \frac  {1}{k!2^{k+1}}\cdot \left( \frac {n-1}{n}\right)^{k+1}.$$ For fixed $k$ we have $\lim_{n\to \infty}(\frac {n-1}{n})^{k+1}=1.$ So $n!/k!n^{k+1}\to \infty.$
Method 2... Less precise than Stirling's formula, but sufficient here:
For integer $A\geq 2$ we have $$\log A=\int_{A-1}^A\log A\;dx>\int_{A-1}^A\log x\;dx.$$ Since $\log 1=0,$  we have, for $n\geq 2$ : $$\log (n!)=\sum_{A=2}^n\log A>\sum_{A=2}^n\int_{A-1}^A\log x\;dx=$$ $$=\int_1^n\log x \;dx=(-x+x\log x)|_1^n=(n\log n)-n+1.$$  So for $n\geq 2$ we have $\log (n!/k!n^{k+1})>n(-k-2+\log n)+1-\log (k!).$
