# simple yet clever proof with factorials

• Oct 20th 2009, 10:29 PM
hairy
simple yet clever proof with factorials
we're supposed to use the principle of mathematical induction for this.
http://dl.getdropbox.com/u/594924/fac.jpg

basically the inductive part boils down to proving algebraically that
$(k+1)!-1+(k+1)\cdot(k+1)!=(k+2)!-1$

but i'm not familiar enough with factorials to show this. i've read about factorials on wikipedia but they didn't have any good identities or anything i could use. any suggestions would be very helpful but for now i'm going to just stare at it for a few minutes and read about series with factorials.

thanks guys
• Oct 20th 2009, 10:45 PM
tonio
Quote:

Originally Posted by hairy
we're supposed to use the principle of mathematical induction for this.
http://dl.getdropbox.com/u/594924/fac.jpg

basically the inductive part boils down to proving algebraically that
$(k+1)!-1+(k+1)\cdot(k+1)!=(k+2)!-1$

but i'm not familiar enough with factorials to show this. i've read about factorials on wikipedia but they didn't have any good identities or anything i could use. any suggestions would be very helpful but for now i'm going to just stare at it for a few minutes and read about series with factorials.

thanks guys

If you already realized that $n!=n(n-1)!$ then it is almost trivial

$(k+1)!-1+(k+1)\cdot(k+1)!=(k+1)!\left[1+k+1\right]-1=(k+1)!(k+2)-1=...$
• Oct 20th 2009, 11:00 PM
hairy
Quote:

Originally Posted by tonio
If you already realized that $n!=n(n-1)!$ then it is almost trivial

$(k+1)!-1+(k+1)\cdot(k+1)!=(k+1)!\left[1+k+1\right]-1=(k+1)!(k+2)-1=...$

thanks, i see.