Question

Give the proof of 0!=1
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Answer 1

Logical proof

Let $n$ be a positive integer. Then, the continued product of first $n$ natural numbers is called factorial. It is denoted by $n!$.

We need to proof $0! = 1$, in that many mathematical formulae work. For example we would like;

$\implies n! = n \times (n - 1)!$

Solving this formula/equation, we get:

$\implies n = \dfrac{n!}{(n- 1)!}$

When $n = 1$ then,

$\implies 1 = \dfrac{1!}{(1 - 1)!} \\ \\ \implies 1 = \dfrac{1!}{0!} \\ \\ \implies 0! = \dfrac{1}{1} \\ \\ \implies \boxed{0! = 1}$

Hence, this proves 0! = 1.

$\rule{90mm}{2pt}$

REMEMBER!

$\boxed{\begin{array}{l} \rm{(i) \: \: n! = n(n - 1)(n - 2) ... 3 \times 2 \times 1.} \\ \\ \rm{(ii) \: \: n! = n \times (n - 1)!.} \\ \\ \rm{(iii) \: \: \rm{0! = 1.}}\end{array}}$

Answer 2

The most appropriate and fundamental method has already been added by @EmpireWarrior, I will use the definition of gamma function to prove the required result.

Solution:

First of all, see the first and second attachments which are the graphs of $x!$ and $\Gamma(x+1)$ respectively.

We can conclude from the graph that $x! = \Gamma(x+1) \, \, \forall x \geqslant 0$

Gamma function can be defined in terms of factorial as follows:

$\boxed{x! = \Gamma(x+1) = \int\limits_{0}^\infty t^x e^{-t} dt}$

We will substitute $x = 0$ to obtain the factorial of 0.

$\implies 0! = \Gamma(1) = \int\limits^\infty_0 t^0 e^{-t} dt$

$\implies 0! = \int\limits^\infty_0 (1) e^{-t} dt$

$\implies 0! = \int\limits^\infty_0 e^{-t} dt$

$\implies 0! = \left |- e^{-t} \right|^ \infty_0$

$\implies 0! = - \left | \dfrac{1}{ {e}^{t} } \right|^ \infty_0$

$\implies 0! = - \left | \dfrac{1}{ {e}^{ \infty} } - \dfrac{1}{ {e}^{0} } \right |$

$\implies 0! = - \left | \dfrac{1}{ \infty} - \dfrac{1}{1 } \right |$

$\implies 0! = - \left | 0 - 1\right |$

$\implies 0! = 1$

Hence it is proved that 0! = 1.

Learn more:

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