combinatorics question chapter 4 number 30 prove for all positive integers $n> k \ge 2$ the inequality $$k^n < \binom{kn}n\;.$$ Bóna third edition, chapter $4$, problem #$30$
 A: HINT: Suppose that you have a matrix with $n$ rows and $k$ columns. There are $k^n$ ways to choose a sequence of $n$ columns. Say that $\langle c_1,c_2,\ldots,c_n\rangle$ is such a sequence; it corresponds to picking the $n$ elements in row $r$, column $c_r$ for $r=1,\ldots,n$. $\binom{kn}n$ is also the number of ways of picking something from the matrix; what?
A: The binomial coefficient is 
$$\frac{(kn)(kn-1)(kn-2)\cdots (kn-n+1)}{(n)(n-1)(n-2)\cdots(1)}.$$
This can be rewritten as
$$\left(\frac{kn}{n}\right)\left(\frac{kn-1}{n-1}\right)\left( \frac{kn-2}{n-2}\right)\cdots \left(\frac{kn-n+1}{1}\right).\tag{1}$$
Expression (1) is a product of $n$ terms.  The first term $\frac{kn}{n}$, by cancellation, is equal to $k$.  We will show that all the other terms are $\gt k$. This will imply that the product of the $n$ terms is $\gt k^n$.
So we want to show that for any $t$ such that  $1\le t \le n-1$, we have 
$\frac{kn-t}{n-t}\gt k$.
Equivalently, we want to show that $kn-t\gt kn-kt$. This is equivalent to $-t\gt -kt$, which is equivalent to $t\lt kt$. That is true if $k\gt 1$.  
