Simplifying Factorials

[burt]

I was watching a Khan Academy video and he said that (n+1)! is the same as (n+1)(n!). Why is this?

 

[MarkFL]

Because:

[MATH]n!\equiv\prod_{k=1}^n(k)[/MATH]

 

[burt]

Because:

[MATH]n!\equiv\prod_{k=1}^n(k)[/MATH]

huh?

 

[MarkFL]

Okay, what if I write:

[MATH]n!\equiv n(n-1)(n-2)\cdots3\cdot2\cdot1[/MATH]
Then we must have:

[MATH](n+1)!\equiv (n+1)n(n-1)\cdots3\cdot2\cdot1[/MATH]
But, isn't this also:

[MATH](n+1)!\equiv (n+1)\cdot n(n-1)(n-2)\cdots3\cdot2\cdot1=(n+1)n![/MATH] ?

 

[burt]

Okay, what if I write:

[MATH]n!\equiv n(n-1)(n-2)\cdots3\cdot2\cdot1[/MATH]
Then we must have:

[MATH](n+1)!\equiv (n+1)n(n-1)\cdots3\cdot2\cdot1[/MATH]
But, isn't this also:

[MATH](n+1)!\equiv (n+1)\cdot n(n-1)(n-2)\cdots3\cdot2\cdot1=(n+1)n![/MATH] ?

Oh, I see! Thanks!
Can you explain the notation you used before, just for curiosity's sake?

 

[MarkFL]

Oh, I see! Thanks!
Can you explain the notation you used before, just for curiosity's sake?

It's simply a more efficient notation for products:

[MATH]\prod_{k=1}^n(a_k)=a_1\cdot a_2\cdot a_3\cdots a_{n-1}\cdot a_{n}[/MATH]

 

[burt]

It's simply a more efficient notation for products:

[MATH]\prod_{k=1}^n(a_k)=a_1\cdot a_2\cdot a_3\cdots a_{n-1}\cdot a_{n}[/MATH]

Thanks!

 

[pka]

I was watching a Khan Academy video and he said that (n+1)! is the same as (n+1)(n!). Why is this?

Khan Academy is part of your problem. Khan takes great liberties with foundations.
The factorial notation is for non-negative integers.
So here is definition: Define \(\displaystyle 0!=1\) then if \(\displaystyle n\in\mathbb{Z}^+\) then \(\displaystyle (n!)=n\cdot (n-1)!\)
That is the axiomatic definition way.