Hints for proving $\frac{1}{m^k} \binom{m}{k} < \frac{1}{n^k} \binom{n}{k}$

Question

I want to prove the following exercise from a textbook:

Let $m,n \in \mathbb{N}$ and $m < n$. Prove directly that for $k = 2, \dots, n$ the inequaity

$$\frac{1}{m^k} \binom{m}{k} < \frac{1}{n^k} \binom{n}{k}$$

holds.

My work so far:

I know that you could probably solve this using induction, but this is not how it's intended to be done. So I want to ask for hints that doesn't involve using induction.

Since $m < n$, it follows $$\forall \alpha < k: m - \alpha < n - \alpha$$

This implies $$\prod_{i = 0}^{k-1} (m - i) < \prod_{i = 0}^{k-1} (n - i)$$

This implies $$\binom{m}{k} = \frac{1}{k!}\prod_{i = 0}^{k-1} (m - i) < \frac{1}{k!}\prod_{i = 0}^{k-1} (n - i) = \binom{n}{k}$$

So, what I've got is $$\binom{m}{k} < \binom{n}{k}$$

Now, I am pretty much stuck. If $m < n$, then $\frac{1}{n} < \frac{1}{m}$, so $$\frac{1}{n^k} < \frac{1}{m^k}$$

which would imply that

$$\frac{1}{n^k} \binom{m}{k} < \frac{1}{m^k} \binom{n}{k}$$

So I guess I've messed something up

Answer 1

Hint:   write it as $\; \displaystyle \frac{1}{k!}\,\prod_{i=0}^{k-1} \,\left(1-\dfrac{i}{m}\right) \;\lt\; \frac{1}{k!}\,\prod_{i=0}^{k-1} \,\left(1-\dfrac{i}{n}\right)\;$ and compare the terms.

Answer 2

$1-\frac j m < 1- \frac j n$ for $j=0,1,2,...,k-1$. Just multiply these inequalities and simplify.