Factorial (0!=1?)

Why \( 0! = 1 \)?

Recursive Definition

The factorial function is defined recursively as
\(n! = n \cdot (n - 1)! \quad \text{for } n \geq 1\)
To maintain consistency, consider \( n = 1 \), then
\(1! = 1 \cdot 0!\)
Since \( 1! = 1 \), it follows that
\(1 = 1 \cdot 0!\)
or \(0! = 1\)

Algebraic / Combinatorial Definition

Factorial \( n! \) counts the number of permutations (arrangements) of \( n \) distinct objects.
For example

1. \( 3! = 6 \) ways to arrange 3 objects
2. \( 1! = 1 \) way to arrange 1 object
3. How many ways to arrange zero objects?

There is exactly one way: the empty arrangement. Hence, by combinatorial reasoning, we get
\(0! = 1\)

Pattern / Numerical Continuation

Observe the factorial pattern by working backward using \( n! = \dfrac{(n+1)!}{n+1} \), we get
\(\begin{align*} 4! &= \frac{5!}{5} = \frac{120}{5} = 24 \\ 3! &= \frac{4!}{4} = \frac{24}{4} = 6 \\ 2! &= \frac{3!}{3} = \frac{6}{3} = 2 \\ 1! &= \frac{2!}{2} = \frac{2}{2} = 1 \\ 0! &= \frac{1!}{1} = \frac{1}{1} = 1 \\ \end{align*}\)
Thus
The natural numerical pattern leads to \( 0! = 1 \).

Gamma Function (Analytic Continuation)

The Gamma function \( \Gamma(z) \) extends the factorial to complex numbers and satisfies that
\(\Gamma(n) = (n - 1)! \quad \text{for positive integers } n\)
Therefore
\(0! = \Gamma(1)\)
By definition of the Gamma function, we get
\(\Gamma(1) = \int_0^\infty t^{1 - 1} e^{-t} \, dt = \int_0^\infty e^{-t} \, dt = \big[ -e^{-t} \big]_0^\infty = 0 - (-1) = 1\)

Summary Table

Approach	Reasoning	Result
Recursive Definition	\( 1! = 1 \cdot 0! \Rightarrow 0! = 1 \)	\( 0! = 1 \)
Combinatorics	One way to arrange zero objects (empty permutation)	\( 0! = 1 \)
Numerical Pattern	Continuing \( n! = \frac{(n+1)!}{n+1} \) down to \( n = 0 \)	\( 0! = 1 \)
Gamma Function	\( 0! = \Gamma(1) = \int_0^\infty e^{-t} dt = 1 \)	\( 0! = 1 \)