# Thread: A proof involving logs

1. ## A proof involving logs

Can anyone show that:

.......................N-1
log(N-1)! = $\Sigma$n=1 log(n)

NB: The dots leading up to the "N-1" on the top line are just to make it line up with the n=1 as it part of the summation bit and i couldnt work out how to do it using latex math!

2. Originally Posted by Mathman87
Can anyone show that:

.......................N-1
log(N-1)! = $\Sigma$n=1 log(n)

NB: The dots leading up to the "N-1" on the top line are just to make it line up with the n=1 as it part of the summation bit and i couldnt work out how to do it using latex math!
$\log(n) + \log(m) = \log(mn)$

So $\sum_{n=1}^{N-1} \log(n) = \log(1) + \log(2) + \log(3) + \dots + \log(N-1)$

$= \log(1\cdot2\cdot3 \dots \cdot (N-1)) = \log((N-1)!)$

3. makes sense, thank you! :-)