Is this a demonstration or a definition? Some people says that this can be demonstrated $0!=1$, but other say that this is a definition. Which one is correct?
Let's given $n\in\mathbb{N}$:
$$(n+1)!=(n+1)\cdot n!$$
$$(0+1)!=(0+1)\cdot 0!$$
$$1!=1\cdot 0!$$
$$1=0!$$
 A: Depending on the equations suitable to define $n!$, there are multiple possible answers:
If $n! := n (n-1)!, 0!=1$ is used, it's a plain definition.
If $1! = 1$ is used as the starting point, it can be proved that $0!:=1$ is a consistent extension. (this is what you did)
If $n! := \prod_{i\in \mathbb N, i \le n} i$, it follows from the definition of the empty product, $\prod_{k\in\emptyset} f(k) := 1$.
If $n! := \Gamma(n+1), n\in\mathbb N$, it follows from the definition because $\Gamma(1) = 1$.
If $n! := |S_n| = |\{f: A\to A \text{ bijective}\}|$ with $|A| = n$, there is only one set with $0$ elements, $\emptyset$ and only one function $f: \emptyset \to\emptyset$
A: There are multiple equivalent ways of defining the factorial function.  Since the exact choice of definition doesn't matter in the long run, people don't generally make a big deal about which definition they're choosing.  In some contexts, this would be a definition, whereas in others it would be a proof from the definition.
A: In demonstrating that $0! = 1$, you assumed that $1! = 1$.
If you define $n!$ by 


*

*$1! = 1$

*$n! = n(n - 1)!$ for $n \in \mathbb{N}$
with $\mathbb{N}$ defined as the set of positive integers, then you can demonstrate that $0! = 1$ by substituting $1$ for $n$.  
If, as some mathematicians do, you define $\mathbb{N}$ to be the set of non-negative integers, it would make sense to define $n!$ by 


*

*$0! = 1$

*$(n + 1)! = (n + 1)n!$ for $n \in \mathbb{N}$, with $n \geq 1$
A: Another way of showing why this should be true:
\begin{align}
5!&=5\cdot4\cdot3\cdot2\cdot1\cdot(1)\\
4!&=4\cdot3\cdot2\cdot1\cdot(1)\\
3!&=3\cdot2\cdot1\cdot(1)\\
2!&=2\cdot1\cdot(1)\\
1!&=1\cdot(1)\\
0!&=(1)
\end{align}
The first five lines are easy to verify, and the last line follows by analogy.
