Why is 0! equal to 1?

The natural guess is that an empty arrangement should count as zero, or as nothing at all. That confuses an arrangement with an item. An arrangement is a way of finishing the job. With no items there is no choice to make and nothing left over, so the job is already finished in exactly one way: do nothing. Calling that zero would say there is no way to have finished arranging nothing, which is false. The empty shelf is a result, not a missing result.

The shelf example makes the distinction concrete. With one book there is one shelf: the book in the only position. Remove the book. The shelf does not disappear; there is still one possible state of the shelf, the empty one. Counting arrangements is counting states, and an empty state is a state. Equivalently, an arrangement assigns each item to a distinct position. For zero items there are no items and no positions to fill, and there is exactly one assignment, the empty assignment. You do not get zero assignments; you get the one assignment that maps nothing to nothing.

A factorial is also a product: $n! = 1\times2\times\cdots\times n$. When $n=0$ the product has no factors. The empty product is $1$ because $1$ is the identity for multiplication: multiplying by $1$ leaves a product unchanged, so starting with $1$ and multiplying by nothing leaves $1$. This is the same fact as the empty arrangement, written in the language of products. It is not a convention bolted on to make formulas tidy. It is what the multiplication rule gives when the pile starts empty and there is one way for it to stay empty.

Set $0!=0$ and the recurrence gives $1!=1\times0!=0$. But $1!$ is the number of orders of one item, and that number is one: the item alone. Continuing, $2!=2\times0=0$, $3!=3\times0=0$, and every higher factorial collapses to zero. The counting formulas at the edges fail too. The number of ways to choose none from five is one, and the formula gives the same value only when $0!=1$. With $0!=0$ that expression divides by zero. The value $0!=1$ is what keeps the same rule working at the boundary.

This is a count for whole numbers of items, zero included. Negative integers do not describe a pile that can be arranged, so ordinary factorial is not defined there; the gamma function extends the formula but is not counting arrangements. If some items are indistinguishable, the count is divided by the arrangements of the identical items, but at zero items there is nothing to divide and the empty arrangement remains one. The value is not saying there is one item. It is saying there is one arrangement of zero items.

Read upward, the recurrence starts at the empty case: $0!=1$, then $1!=1\times1=1$, $2!=2\times1=2$, $3!=3\times2=6$, $4!=4\times6=24$, $5!=5\times24=120$. The chain has to begin somewhere. The empty arrangement is the only start that makes the first step correct. If you begin at $0!=0$, the first step gives $1!=0$, and no later step can repair it because multiplying zero by anything remains zero.

Zero factorial is not a statement that there is one item. It is a statement about the number of arrangements of an empty collection. The collection has size zero; the count of its arrangements has size one. Those are different numbers, and there is no contradiction in them being different. A set with no elements has one subset, namely itself; similarly it has one ordering, namely the empty ordering. That does not make zero into one; it makes the count of orderings of nothing equal to one.