Algebraically solve nC7 = n+1C8?

nC7 = n! / 7!(n-7)!
n+1C8 = (n+1)! / 8!(n+1-8)!

nC7 = n+1C8

Rearranging gives:
nC7 - n+1C8 = 0

Rewrite in factorial form
n! / 7!(n-7)! - (n+1)! / 8!(n+1-8)! = 0

Rewrite 8! in the denominator
n! / 7!(n-7)! - (n+1)! / 7!*8(n-7)! = 0

Rewrite with common denominator
n!*8 / 7!*8(n-7)! - (n+1)! / 7!*8(n-7)! = 0

Divide by n! / 7!*8(n-7)!
8 - (n+1) = 0

8 - n - 1 = 0

7 - n = 0

n = 7