How to show that $\dfrac {n!}{(n-k)!} \leq n^{k}$

Question

I wish to show the following in equality:

$$\dfrac {n!}{(n-k)!} \leq n^{k}$$

Attempt:

$$\dfrac {n!}{(n-k)!} = \prod\limits_{i = 0}^{k -1} (n-i) = n\times (n-1)\times \cdots \times(n-({k-1})) $$

I am not sure how to make the argument that $n\times (n-1)\cdots \times (n-({k-1})) \leq n^k$

Answer 1

Each of the $k$ terms in the product is less than or equal to $n$.

ALternatively, the LHS counts the number of injective maps from a set with $k$ elements to a set with $n$ elements while the RHS counts all the maps from a set with $k$ elements to a set with $n$ elements. The result follows.

Answer 2

If $n-i\le n,$ then the product of $k$ of those is clearly $\le n^k.$

Answer 3

\begin{eqnarray*} \underbrace{n(n-1) \cdots (n-k+1)}_{ k \text{ terms}} \underbrace{ \leq}_{ n-i \leq n} n^k. \end{eqnarray*}