Why Zero Factorial is Equal to One

Why should you be concerned that 0!=1? You may not even know what a factorial number is.

Reading the following you may come to understand the idea of a factorial.You may also be able to please your friends and confound your enemies bybeing able to show that 0!=1. Here is an explanation, requiring only knowledge of simple arithmetic tounderstand.

When considering the numbers of different groups which may be formed from acollection of objects one frequently finds the need to make calculations ofthe form, 4 x 3 x 2 x 1 for 4 objects. I suppose I must explain why that is.

Say we have the four objects A, B, C, D.
If I select from that collection to form another group, you will see that Icould choose any one of the four for the first selection. Thiswould leave three objects. I could select any of those three for the second selection,leaving any one of two for the next and so on. The total number of ways I could selectthose objects would be 4 x 3 x 2 x ... , and the series would stop at one when I selectedthe last object. To save repeatedly writing down long strings of such products thenotation 4! is used to represent 4 x 3 x 2 x 1 or 6! for 6 x 5 x 4 x 3 x 2 x 1.The ! is read as factorial. So the examples quoted above are more easily written as 4!and 6!. If you care to calculate their values they are 24 and 720 respectively.

In general any factorial number (call it n!), may be written,
n! = n x (n-1) x (n-2) x (n-3) x ... x 2 x 1
This is the general definition of a factorial number.
If you want it in words; a factorial number is the product of all positive integers from1 to the number under consideration.
The main place it is likely to be encountered is when considering those groups andarrangements of objects mentioned above.

So where does all this 0! stuff fit in?

Nobody has trouble in stating 2! = 2 x 1 , or even that 1! = 1, but 0! appears to make nosense.
It does however, have a value of 1. This is rather counter intuitive but arisesdirectly from our general definition.
n! = n x (n-1) x (n-2) x (n-3) x ... x 2 x 1

Notice this may be written,
n! = n x (n-1)! Still exactly the same definition.

If the left hand side (LHS) = the right hand side (RHS) then dividing both sides by nshould leave them still equal, so it is still true to write,
n!/n = n x (n-1)!/n

The (n-1)! in the RHS is being both multiplied by n and divided by n. These cancelleaving,
n!/n = (n-1)! If you doubt this, try it with real numbers, e.g. 4!/4 = 3! or (4 x 3 x 2 x 1)/4 = 3 x 2 x 1 = 6

The equation we now have is,
n!/n = (n-1)!
It is still our original definition in arearranged form. For convenience I shall write it the other way round.
(n-1)! = n!/n

We also said that our factorial uses the positive integers 1 and above.Try the value of n=2 in our rearranged formula and we get,
(2-1)! = 2!/2 or 1! = 2x1/2
The RHS calculates to 1, so we have the statement 1!=1That is what we guessed intuitively above. It is now confirmed.But look what happens when we substitute the legitimate value of n=1 in our formula.
(1-1)! = 1!/1
Evaluating this statement gives
0! = 1!/1
We have just shown 1!=1 so the RHS is 1/1 or 1.

Our rearranged definition of a factorial number gives directly the statement,
0!=1 Counter intuitive perhaps but if the definition is true then this is true.

Why not tell your friends about it? Help dispel the widespread ignorance about 0!

ADDENDUM
I feel obliged to add this comment to the above. From time to time I get people writing to me to tell me that I am dividing by zero and that this is negating my 'proof'. First let me say that is not a proof, it is merely a restatement of the definition of a factorial number. Secondly, all I have done is to rearrange the definition in order to get a meaning for (n-1)!. This shows that when n=1 (a perfectly valid value for n), the expression has a value of 1. There is never a substitution of zero for n at any time. Substituting n=1, (NOT ZERO), shows that (1-1)! = 1. Factorial numbers are ALWAYS positive NON-ZERO integers. It is just that in computations a value is sometimes required for (1-1)!. I am showing why that value is 1. It is an arbitrary convention that (1-1)! has become known as 0!.