Finding value of $\lim\limits_{n\rightarrow \infty}\Big(\frac{(kn)!}{n^{kn}}\Big)^{\frac{1}{n}}$ 
Finding value of $\displaystyle \lim_{n\rightarrow \infty}\bigg(\frac{(kn)!}{n^{kn}}\bigg)^{\frac{1}{n}}$ for all $k>1$

Try: I have solved it using stirling Approximation
$\displaystyle n!\approx \bigg(\frac{n}{e}\bigg)^n\sqrt{2\pi n}$ for laege $n$
So we have $\displaystyle \lim_{n\rightarrow \infty}\bigg(\frac{kn}{e}\bigg)^{kn}\cdot \bigg(\sqrt{2\pi k n}\bigg)^{\frac{1}{n}}\cdot \frac{1}{n^k}=\bigg(\frac{k}{e}\bigg)^k$
Could some help me how to solve it without 
stirling Approximation
Thanks.
 A: Simply note that this is a Riemann sum for positive $k$.
Letting $$\lim_{n\to\infty}\left(\frac{(kn!)}{n^{kn}}\right)^{1/n}=y\,,$$ we have
\begin{align}
 \ln y&=\lim_{n\to\infty}\frac{1}{n}\ln\left(\frac{(kn)!}{n^{kn}}\right)
\\&=\lim_{n\to\infty}\frac{1}{n}\ln\left(\prod_{j=1}^{kn}\frac{j}n\right)
\\&=\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{kn}\ln\left(\frac{j}n\right)
\\&=\int_0^{k}\ln x\,dx=\ln\left(\frac{k}e\right)^k
\end{align}
We can interchange log and limit as $\ln(\cdot)$ is continuous in its domain.
A: $$  \lim_{n\rightarrow \infty} \left(\dfrac{(kn)!}{n^{kn}} \right)^{\dfrac{1}{n}}
= \lim_{n\rightarrow \infty} \left(\dfrac{(kn)!}{\dfrac {kn^{kn}}{k}} \right)^{\dfrac{k}{kn}}
= {(\lim_{n\rightarrow \infty} \left(\dfrac{(kn)!}{\dfrac {kn^{kn}}{k}} \right)^{\dfrac{1}{kn}}})^{k}
= \left(\dfrac{k}{e} \right)^k $$
The last step is derived from: 
$$y =\lim_{n\rightarrow \infty}\bigg(\dfrac{(n)!}{n^{n}}\bigg)^{\dfrac{1}{n}}$$
$\ln{y} =\displaystyle \lim_{n\rightarrow \infty} 1/n × \ln(1/n) × \ln(2/n) × ... × \ln(1)$ 
= $\sum_{r=1}^{n}1/n × \ln{r/n}$
=  $\int_0^{1}\ln x\,dx$ 
= $-1$ 
(Converting infinite sum into an integral) 
And therefore, $y = e^{-1}$. Plug in the same above. 
