Show that $r_k^n/n \le \binom{kn}{n} < r_k^n$ where $r_k = \dfrac{k^k}{(k-1)^{k-1}}$ Show that 
for $n \ge 2$, 
$\dfrac{r_k^n}{n+1} \le \binom{kn}{n} < r_k^n$ where $r_k = \frac{k^k}{(k-1)^{k-1}}$.
This is a generalization of 
How to prove through induction
which asks for a proof that
$\binom{2n}n<4^n$.
I wondered what was true for
$\binom{kn}{n}$
and this is what I came up with.
Note that
$r_k
=k(1+\frac1{k-1})^{k-1}
\sim ek
$
for large $k$.
The key is to get
 bounds for
$\dfrac{\binom{k(n+1)}{n+1}}{\binom{kn}{n}}$
that are not too dependent on $n$.
The upper bound I get is not,
the lower bound is,
but telescopes nicely.
Note:
The algebra here
is uncomfortably messy.
I would definitely like
a simpler proof,
perhaps depending on
Stirling's theorem.
Here goes.
$\begin{array}\\
s_k(n)
&=\dfrac{\binom{k(n+1)}{n+1}}{\binom{kn}{n}}\\
&=\dfrac{\frac{(kn+k)!}{(n+1)!(kn+k-n-1)!}}{\frac{(kn)!}{n!(kn-n)!}}\\
&=\dfrac{n!(kn-n)!(kn+k)!}{(n+1)!(kn)!(kn+k-n-1)!}\\
&=\dfrac{\prod_{j=0}^{n}(kn+k-j)}{(n+1)\prod_{j=0}^{n-1}(kn-j)}\\
&=\dfrac{\prod_{j=0}^{k-1}(kn+k-j)\prod_{j=k}^{n}(kn+k-j)}{(n+1)\prod_{j=0}^{n-1}(kn-j)}\\
&=\dfrac{\prod_{j=0}^{k-1}(kn+k-j)\prod_{j=0}^{n-k}(kn-j)}{(n+1)\prod_{j=0}^{n-1}(kn-j)}\\
&=\dfrac{\prod_{j=0}^{k-1}(kn+k-j)}{(n+1)\prod_{j=n-k+1}^{n-1}(kn-j)}\\
&=\dfrac{\prod_{j=0}^{k-1}(kn+k-j)}{(n+1)\prod_{j=0}^{k-2}(kn-(j+n-k+1)}\\
&=\dfrac{\prod_{j=0}^{k-1}(kn+k-j)}{(n+1)\prod_{j=0}^{k-2}((k-1)n-(j-k+1)}\\
&=\dfrac{(kn+k)\prod_{j=1}^{k-1}(kn+k-j)}{(n+1)\prod_{j=0}^{k-2}((k-1)n+k-j-1)}\\
&=k\dfrac{\prod_{j=0}^{k-2}(kn+k-j-1)}{\prod_{j=0}^{k-2}((k-1)n+k-j-1)}\\
&=k\prod_{j=0}^{k-2}\dfrac{kn+k-j-1}{((k-1)n+k-j-1)}\\
&<k\left(\dfrac{k}{k-1}\right)^{k-1}\\
&=\dfrac{k^k}{(k-1)^{k-1}}\\
\end{array}
$
since,
if $1 \le a \le k-1$,
$\begin{array}\\
\dfrac{kn+a}{(k-1)n+a}-\dfrac{k}{k-1}
&=\dfrac{(kn+a)(k-1)-k((k-1)n+a)}{(k-1)((k-1)n+a)}\\
&=\dfrac{(k^2n-kn+ka-a)-(k^2n-kn+ak)}{(k-1)((k-1)n+a)}\\
&=\dfrac{-a}{(k-1)((k-1)n+a)}\\
&< 0\\
\end{array}
$
To  get the lower bound,
$\begin{array}\\
\dfrac{kn+a}{(k-1)n+a}-\dfrac{k}{k-1}
&=\dfrac{-a}{(k-1)((k-1)n+a)}\\
&=\dfrac{-1}{(k-1)((k-1)n/a+1)}\\
&\ge\dfrac{-1}{(k-1)(n+1)}\\
&=-\dfrac{1}{(k-1)n+1}\\
\end{array}
$
or
$\begin{array}\\
\dfrac{kn+a}{(k-1)n+a}
&\ge\dfrac{k}{k-1}-\dfrac{1}{(k-1)n+1}\\
&=\dfrac{k((k-1)n+1)-(k-1)}{(k-1)((k-1)n+1)}\\
&=\dfrac{k(k-1)n+1}{(k-1)((k-1)n+1)}\\
&=\dfrac{k(k-1)n+1}{(k-1)^2n+k-1}\\
&>\dfrac{k(k-1)n}{(k-1)^2n+k-1}\\
&=\dfrac{kn}{(k-1)n+1}\\
&=\dfrac{k}{(k-1)+1/n}\\
&=\dfrac{k}{(k-1)(1+1/(n(k-1))}\\
&=\dfrac{k}{k-1}\dfrac1{1+1/(n(k-1))}\\
\end{array}
$
so that
$s_k(n)
\ge k\prod_{j=0}^{k-2}\dfrac{k}{k-1}\dfrac1{1+1/(n(k-1))}
= \dfrac{k^k}{(k-1)^{k-1}}\prod_{j=0}^{k-2}\dfrac1{1+1/(n(k-1))}
= \dfrac{k^k}{(k-1)^{k-1}}\dfrac1{(1+1/(n(k-1)))^{k-1}}
$.
In an earlier problem
I showed 
(here: Prove that, if $0 < x < 1$, then $(1+\frac{x}{n})^n < \frac1{1-x}$)
that
if $0 < x < 1$
then
$(1+x/n)^n < \frac1{1-x}$.
Therefore
$(1+1/(n(k-1)))^{k-1}
<\frac1{1-1/n}
$
so that
$s_k(n)
>\dfrac{k^k}{(k-1)^{k-1}}(1-1/n)
=\dfrac{k^k}{(k-1)^{k-1}}\dfrac{n-1}{n}
=r_k\dfrac{n-1}{n}
$.
The products of
$\dfrac{n-1}{n}$
telescope
to give the lower bound stated.
 A: No one else has done anything,
so I thought I'd see what happens
when I follow my own suggestion
of using Stirling's formula.
Definitely easier
and, of course,
more precise.
Here we go.
Since
$n! \approx \sqrt{2\pi n}(n/e)^n$,
$\begin{array}\\
\binom{kn}{n}
&=\dfrac{(kn)!}{n!(kn-n)!}\\
&\sim \dfrac{\sqrt{2\pi kn}(kn/e)^{kn}}
{(\sqrt{2\pi n}(n/e)^n)(\sqrt{2\pi (kn-n)}((kn-n)/e)^{kn-n})}\\
&= \sqrt{\dfrac{2\pi kn}{(2\pi n)(2\pi (kn-n))}}\dfrac{(kn)^{kn}}
{((n)^n)(((kn-n))^{kn-n})}\\
&= \sqrt{\dfrac{k}{2\pi n(k-1)}}\dfrac{k^{kn}n^{kn}}
{(n)^n((n(k-1))^{nk-n)})}\\
&= \sqrt{\dfrac{k}{2\pi n(k-1)}}\dfrac{k^{kn}}
{(((k-1))^{kn-n})}\\
&= k^n\sqrt{\dfrac{k}{2\pi n(k-1)}}\left(\dfrac{k}
{k-1}\right)^{n(k-1)}\\
&= k^n\sqrt{\dfrac{k}{2\pi n(k-1)}}\left(\dfrac{k}
{k-1}\right)^{kn-n}\\
&= \sqrt{\dfrac{k}{2\pi n(k-1)}}k^n\left(\left(\dfrac{k}
{k-1}\right)^{k-1}\right)^{n}\\
&= \sqrt{\dfrac{k}{2\pi n(k-1)}}\left(\dfrac{k^k}
{(k-1)^{k-1}}\right)^{n}\\
\end{array}
$
This confirms my main result
of the ratio of consecutive
terms being
$\dfrac{k^k}
{(k-1)^{k-1}}
$,
with the multiplier of
$\sqrt{\dfrac{k}{2\pi n(k-1)}}$
being nicely between
$\dfrac1{n}$ and $1$.
