Why is 0! equal to 1?

Great question! The value of \(0! = 1\) might seem a bit counterintuitive at first, but it is well justified both mathematically and logically. From a **combinatorial perspective**, the factorial \(n!\) represents the number of ways to arrange \(n\) distinct objects. For example, \(3! = 6\) because there are 6 ways to arrange 3 objects. When \(n = 0\), the question becomes: "How many ways are there to arrange zero objects?" There is exactly one way to do nothing — the empty arrangement. So, by definition, \(0! = 1\). From an **algebraic perspective**, the factorial function can be defined recursively: \[ n! = n \times (n-1)! \] If we apply this at \(n=1\): \[ 1! = 1 \times 0! \] Since \(1! = 1\), it follows that: \[ 1 = 1 \times 0! \implies 0! = 1 \] This recursive definition requires \(0!\) to be 1 to keep the factorial function consistent. Additionally, from the perspective of the **gamma function**, which generalizes factorials to real and complex numbers, \(\Gamma(n) = (n-1)!\) for positive integers \(n\). The gamma function is defined such that: \[ \Gamma(1) = \int_0^\infty e^{-t} t^{1-1} dt = \int_0^\infty e^{-t} dt = 1 \] Therefore, \[ 0! = \Gamma(1) = 1 \] In summary, defining \(0! = 1\) maintains consistency in combinatorics, recursive formulas, and continuous extensions of factorials, making it a natural and necessary convention in mathematics.