Consider a sequence of $m$ empty boxes
\begin{equation}
\begin{split}
f(1) & = \square \\
f(2) & = \square \\
f(3) & = \square \\
\quad & \enspace \vdots \\
f(m) & = \square
\end{split}
\end{equation}
and the task of allocating the integers $1, \ldots, n$ to the boxes such that all boxes contain an integer, no box contains more than one integer, and that each integer is used only once. If we start allocating the numbers one by one, then for the first box we have $n$ integers to choose from. Having filled the first box, for the second box there are $n - 1$ choices to choose from. Continuing in this way, we find that there are
\begin{equation}
n (n - 1) (n - 2) \cdots (n - m + 1)
= \frac{n!}{(n - m) !}
\end{equation}
distinct ways of filling the $m$ boxes with $n$ integers. Each allocation defines an injection, and each injection $\{ 1, \ldots, m \} \to \{ 1, \ldots, n \}$ corresponds to some such allocation, so we are done. $\square$
