the base case of P(7) is true because of the computation, where 3^7 < 7!
now in the induction step we assume its true for some arbitrary value n,
3^n < n! and here we have to show it is also true for n+1
3^n+1 = 3 x 3^n < 3 x n! (as we assumed P(n) is true for n >= 7)
3 x n! < (n + 1) x n! because we only talk about values of n >= 7
and (n+1) x n! is simply (n+1)!
and we arrive at the statement 3^n+1 < (n+1)! so it holds and is true for all n >= 7 :)
#proove by mathematical induction 2
solid sun
(n+1)! is also the same as (n+1)n! as its simply the definition of the factorial function
since n! = (n)(n-1) .. (1)
and (n+1)! = (n+1)(n)(n-1) . . . (1)