Proof using induction: $n! > n^2$, for $n\geq4$

Question

Proof using induction: $n! > n^2$, for $n\geq4$

• Basis: For n = 4, we have: $4! > 4^2$
  $24 > 16$ (TRUE)

• Inductive step:
  By the induction hypothesis:
  $k! > k^2$
  $(k+1)k! > (k+1)k^2$
  $(k+1)! > k^3 + k^2$
  But, $k^3 + k^2 > k^2 + 2k + 1$, for $k\geq4$

So, $(k+1)! > k^3 + k^2 > k^2 + 2k + 1$
$(k+1)! > (k+1)^2$ ---> What we want to proof

Does this serve as a proof for my sentence?

Answer 1

@Vinicius No need to make that so complicated.

Assuming $k! > k^2$. $$(k+1)!>(k+1)^2$$ $$(k+1)k! > (k+1)(k+1)$$ $$k!>k+1$$

The last statement requires little proof for $k \in \{4, 5, 6, \cdots\}$.

Answer 2

Alternatively lesser steps. We want to prove $(k+1)!>(k+1)^2$ ie proving $k!>(k+1)$ which is by observation or gamma function true thats all.