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^{2}< n! for n > 3

this is what i have done:

the base case (n = 4) is obviously true since 4

^{2}< 4!

now, assume that it is true for n = k, i.e., k

^{2}< k!

now i have to prove it for n = k+1

since k > 3,

1 < k-1

1(k+1) < (k-1)(k+1)

k+1 < k

^{2}- 1 < k

^{2}< k!

k+1 < k!

(k+1)(k+1) < (k+1)k!

(k+1)

^{2}< (k+1)!

what i have done seems ok to me. but is there any simpler way to do the induction step? what i have done seems a bit "forced" (if you know what i mean).

thanks in advance.