Why is $0! = 1$ the same as $1! = 1$? I want to ask why is $$0! = 1$$ the same as $$1! = 1.$$ 
As a student I was lost and when I tried to ask the question the teacher said this will be done in complex analysis.
I know here I will thirst my quest by the help of people who understand it. 
I know We define $n!$ as the product of all integers k with $1\leq k\leq n$. When $n=0$ this product is empty so it should be 1. 
I was just wondering if for factorial we take:
$$6! = 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$$
$$5! = 5\cdot 4\cdot 3\cdot 2\cdot 1$$
In normal multiplication any number multiplied by $0$ = $0$
hence $$0! = 0\cdot 0 = 0$$
hence $$1! = 1\cdot 1 = 1$$
I am confused by this help Thanks.
But then the factorial rule of $0!$ being $1$ is misleading because $0$ x emptyset = emptyset why would it be 1?  
 A: Note that with the usual definition, $n!=n(n-1)(n-2)\cdots1$, we have $n!=n(n-1)!$
If we  extend this using $n=1$, we get $1!=1\cdot0!$.
It is also useful to define a product of $0$ numbers to be $1$. This also works for $0!$ which would be a product of $0$ integers.
The Binomial Theorem also works when we define $0!=1$, for example
$$
\binom{n}{0}=\frac{n!}{(n-0)!\,0!}=1
$$
A: You should watch this Numerphile video on Youtube: 
https://www.youtube.com/watch?v=Mfk_L4Nx2ZI
Here the argument is made that $0!$ should be one because then the rule
$$
n! = \frac{(n+1)!}{n+1}
$$
is true even for $n=0$. One can make other arguments, this is just one.
A: You have written
$$6! = 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\\
5! = 5\cdot 4\cdot 3\cdot 2\cdot 1$$
and so on, but then concluded
$$0! = 0 \cdot 0.$$
This doesn't fit the pattern, which should continue as follows:
$$4! = 4\cdot 3\cdot 2\cdot 1\\
3! = 3\cdot 2\cdot 1\\
2! = 2\cdot 1\\
1! = 1\\
0! = $$
At this point you simply need to define what $0!$ should mean, and it is natural to choose $1$, the multiplicative identity.
A: We have the gamma function that is defined for all complex number $z$, excluding the non-positive integers, specifically $$\Gamma(z) = \int_0^{\infty}t^{z-1}e^{-t} \, \mathrm{d}t$$
We can then define the factorial as $$n! = \Gamma(n+1).$$ For $n=0$, we have $$0! = \Gamma(1) = \int_0^{\infty} e^{-t} \, \mathrm{d}t = 1$$
Which is an improper integral that converges to $1$. So we have $0! = 1$.

It is quite easy to see that the integral converges to $1$, we need only simply evaluate the definite integral as follows $$\Gamma(1) = \lim_{a \to \infty} \int_0^{a} e^{-t} \, \mathrm{d}t = \lim_{a \to \infty}\left[-e^{-t}\right]_0^a = \lim_{a\to \infty} \left(1 - e^{-a}\right) = 1.$$
A: We have by definition $(n+1)! = (n+1)\,n!$. Therefore $1! = (0+1)! = (0+1)\,0! = 0!$
A: The number $n!$ counts the number of bijections from a set of $n$ elements to itself. There is exactly one bijection from the emptyset to itself, the empty function. Thus $0!=1$.
A: You can think of the definition of the factorial backwards: 
$$(n-1)!=\frac{n!}{n}.$$
Then you get
\begin{align}
2!&=\frac{1\cdot 2\cdot 3}{3}=1\cdot 2,\\
1!&=\frac{1\cdot 2}{2}=1,\\
0!&=\frac{1}{1}=1.
\end{align}
Voilà!
A: A power of exponent $0$ and positive base is $1$. This means that a product of no factors is $1$
