Equivalence of Definitions of Factorial

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\begin{cases} 1 & : n = 0 \\ n \left({n - 1}\right)! & : n > 0 \end{cases} \equiv \displaystyle \prod_{k \mathop = 1}^n k$

Basis for the Induction
$P \left({0}\right)$ is the case:
 * $\displaystyle \prod_{k \mathop = 1}^0 k = 1$

which holds by definition of vacuous product.

Thus $P \left({0}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 0$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\begin{cases} 1 & : r = 0 \\ r \left({r - 1}\right)! & : r > 0 \end{cases} \equiv \displaystyle \prod_{k \mathop = 1}^r k$

from which it is to be shown that:
 * $\begin{cases} 1 & : r = 0 \\ \left({r + 1}\right) r! & : r > 0 \end{cases} \equiv \displaystyle \prod_{k \mathop = 1}^{r + 1} k$

Induction Step
This is the induction step:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\begin{cases} 1 & : n = 0 \\ n \left({n - 1}\right)! & : n > 0 \end{cases} \equiv \displaystyle \prod_{k \mathop = 1}^n k$