recursion - Complexity computation of a determinant recursive algorithm -

i have written algorithm compute determinant of n x n matrix, base on laplace expansion:

i got recurrence relation below: t(n) = n(n² + t(n-1))

i read in wikipedia should yield result t(n) = o(n!), don't know how prove (although it's intuitive).

the formula correct recurrence relation not. not need n², since don't have save or generate sub matrices.

mij determinant of (n-1) x (n-1) sub matrix. have compute n determinants of n different matrices. correct recurrence relation t(n) = n⋅t(n-1) + 2n-1. simplifies

t(n) = n ⋅ t(n-1) + 2n-1      = n ⋅ (n-1) ⋅ t(n-2) + n ⋅ (n-1)      = n ⋅ ( (n-1) ⋅ ( (n-2) ⋅ (...) + n-3 ) + n-2 ) + n-1      = 2n-1 + n ⋅ (2(n-1)-1) + n ⋅ (n-1) ⋅ (2(n-2)-1) + ... + n!      < 2 ⋅ n + 2 ⋅ n ⋅ (n-1) + 2 ⋅ n ⋅ (n-1) ⋅ (n-2) + ... + 2 ⋅ n! + n!      = 2 ⋅ (n + n ⋅ (n-1) + ... + n!/2) + 3 ⋅ n!      < 2 ⋅ (n!/(n-1)! + n!/(n-2)! + ... + n!/2!) + 3 ⋅ n! 

due 2⋅n!/k! ≤ n!/(k-1)! k ≥ 2, get

  n!/(n-1)! + n!/(n-2)! + n!/(n-3)! + ... + n!/2! ≤ n!/(n-2)! + n!/(n-2)! + n!/(n-3)! + ... + n!/2! ≤ n!/(n-3)! + n!/(n-3)! + ... + n!/2! ≤ n!/(n-4)! + ... + n!/2! ≤ ... ≤ n!/2! + n!/2! ≤ n! 

and so

t(n) = n ⋅ t(n-1) + 2n-1      < 2 ⋅ (n!/(n-1)! + n!/(n-2)! + ... + n!/2!) + 3 ⋅ n!      ≤ 2 ⋅ n! + 3 ⋅ n!      = 5 ⋅ n!      = o(n!) 

so algorithm runs in o(n!)