Explanation:
(n+1)! = (n+1)n(n-1)(n-2)...×3×2
16n! = 16n(n-1)(n-2)...×3×2
16(n-1)! = 16(n-1)(n-2)...×3×2
we can now divide both sides by (n-1)! and get
(n+1)n = 16n + 16
n² + n = 16n + 16
n² - 15n - 16 = 0
1×n² - 15n - 16 = 0
we try to bring this into a form
(n + a)(n - b)
one factor has to contain "-", because we have "-16".
and because -15 = 1 - 16, we see that
a = 1
b = 16
n² - 15n - 16 = (n + 1)(n - 16)
and that is 0, when at least one of the factors is 0, because 0×... = 0.
so, the solutions are
n = -1
n = 16
n = -1 does not make any sense for n!
so, n = 16 is the solution.