As part of homework I'm trying to find the limit of $\sqrt[n]{{kn \choose n}}$ (with $k\in\mathbb{N}$ a given parameter).

I've seen This limit: $\lim_{n \rightarrow \infty} \sqrt [n] {nk \choose n}$., on which the only answer suggests using Stirling's approximation, which I've never learned and I can't seem to understand.

Trying to solve this myself, I first tried finding the ratio directly, which gave me $$\frac{{kn+k \choose n+1}}{{kn \choose n}}=\frac{\frac{\left(kn+k\right)!}{\left(n+1\right)!\left(kn+k-n-1\right)!}}{\frac{kn!}{n!\left(kn-n\right)!}}=\frac{\left(kn+k\right)!n!\left(kn-n\right)!}{\left(n+1\right)!\left(kn+k-n-1\right)!kn!} =\frac{\left(kn+k-n\right)\left(kn+k-n+1\right)\cdots\left(kn+k\right)}{\left(n+1\right)\left(kn-n+1\right)\cdots\left(kn-n\right)} $$ That I'm not really sure how to use.

So instead I tried finding something that I might be able to say is bigger and smaller to use squeeze on, which gave me: $${kn \choose n}=\frac{kn!}{n!\left(kn-n\right)!}=\frac{\left(kn-n+1\right)\cdots\left(kn\right)}{n!}\leq k^{n}\frac{n^{n}}{n!}$$ And then showing that $$ \frac{\frac{\left(n+1\right)^{n+1}}{\left(n+1\right)!}}{\frac{n^{n}}{n!}}=\frac{\left(n+1\right)^{n+1}n!}{n^{n}\left(n+1\right)!}=\frac{\left(n+1\right)^{n}\left(n+1\right)}{n^{n}\left(n+1\right)}=\left(1+\frac{1}{n}\right)^{n}\to\epsilon$$ gives the larger limit of

$$\sqrt[n]{{kn \choose n}}\leq\sqrt[n]{k^{n}\frac{n^{n}}{n!}}=\sqrt[n]{k^{n}}\sqrt[n]{\frac{n^{n}}{n!}}=k\cdot\sqrt[n]{\frac{n^{n}}{n!}}\to k\cdot e $$

But it's obviously not the limit itself since for i.e. $k=1$ the limit is 1 and $k=2$ the limit is 4.

So I'm pretty much stuck, any hints on how I might be able to solve it?