Demonstrating that 1! is = 1

Question

The problem with this explanation is that it's using n = 2 instead of n = 1. Please read the explanation I found on "Math Forum - Ask Dr. Math" ( Link ). How do you do the combination formula when there's only 1 person trying to shake hands?

"In the same way, when thinking about combinations we can derive a formula for "the number of ways of choosing k things from a collection of n things." The formula to count out such problems is n!/k!(n-k)!.
For example, the number of handshakes that occur when everybody in a group of 5 people shakes hands can be computed using n = 5 (five people) and k = 2 (2 people per handshake) in this formula. (So the answer is 5!/(2! 3!) = 10).

Now suppose that there are 2 people and "everybody shakes hands with everybody else." Obviously there is only one handshake. But what happens if we put n = 2 (2 people) and k = 2 (2 people per handshake) in the formula? We get 2! / (2! 0!). This is 2/(2 x), where x is the value of 0!. The fraction reduces to 1/x, which must equal 1 since there is only 1 handshake. The only value of 0! that makes sense here is 0! = 1.

And so we define 0! = 1."

This makes perfect sense, but n = 2. When you try to put 1 / ???? ( 1 - ???? ) it doesn't explain it. Is there a better way to explain 1 factorial?

Answer 1

This answer may be a bit dodgy, but is something worth nothing I think.

You may be used to thinking of factorials as sending an integer to another integer. However, there is a function we have defined that given integer elements, is identical to the factorial function, but maps real numbers to real numbers. Such a function is typically known as a Gamma function, and it is easy to see from its definition that $$1! = \Gamma{}(2) = 1$$

https://en.wikipedia.org/wiki/Gamma_function#Main_definition

Answer 2

We define the factorial of a positive integer $n$ as the product of all positive integers less than or equal to $n$; i.e., $$n! = \prod_{k=1}^n k.$$ This should be consistent with your previous understanding of the notation $n!$. Now so defined, we trivially deduce $1! = 1$. But this is not what you are really asking about; you are asking how the given definition results in $0! = 1$, since it is not immediately obvious how we would define a product of no terms. The product of one term is easy enough to understand. But what is an "empty product?"

We won't directly answer this question (although it does have an answer). Instead, we will motivate the value of $0!$ by the following argument. Note that from the above definition, it is natural that the factorial function should obey the recursion relation $$n! = n(n-1)!$$ Consequently, if we are to attribute a consistent meaning to $0!$, we would substitute $n = 1$ in the recursion relation to obtain $$1! = 1(1-1)! = 0!$$ and since $1! = 1$, we must have $0! = 1$. Also notice that if we try to extend this further, e.g., $$(-1)! = (0 - 1)! = \frac{0!}{0} = \frac{1}{0},$$ we find that $(-1)!$ is not well-defined. And in fact, this is true--there is no analytic continuation of the factorial function that results in a well-defined value of the factorial of negative integers (but that's beyond the scope of your question).

Answer 3

We define that $0! = 1$, and for $n > 0$, that $n! = n \cdot (n-1)!$. This means $1! = 1 \cdot 0! = 1$.

For a combinatorial interpretation, $1!$ is the number of ways to create an ordered list containing a single object, so $1!=1$.