Equivalence of Definitions of Factorial

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Theorem

The following definitions of the concept of Factorial are equivalent:

Definition 1

The factorial of $n$ is defined inductively as:

$n! = \begin{cases} 1 & : n = 0 \\ n \left({n - 1}\right)! & : n > 0 \end{cases}$

Definition 2

The factorial of $n$ is defined as:

\(\displaystyle n!\) \(=\) \(\displaystyle \prod_{k \mathop = 1}^n k\)
\(\displaystyle \) \(=\) \(\displaystyle 1 \times 2 \times \cdots \times \paren {n - 1} \times n\)

where $\displaystyle \prod$ denotes product notation.


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:


\(\displaystyle \prod_{k \mathop = 1}^{r + 1} k\) \(=\) \(\displaystyle \left({r + 1}\right) \prod_{k \mathop = 1}^r k\)
\(\displaystyle \) \(=\) \(\displaystyle \left({r + 1}\right) r!\) Induction Hypothesis

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


Therefore:

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

$\blacksquare$