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We can prove
\[n! \gt 6n^2\]
  for  
\[n \ge 6\]
  using induction.
Let  
\[P(k)\]
  be the statement that the above statement is true. Then
\[6!=720 \gt 216= 6 \times 6"\]

Hence  
\[P(6)\]
  is true.
Suppose that  
\[P(k)\]
  is true.(
\[P(k)\]
  implies  
\[P(k+1)\]
  is true.)
\[\begin{equation} \begin{aligned} (k+1)! &= (k+1) \times k! \\ & \gt (k+1) \times 6k^2 \\ & \gt (k_1) \times 6(k+1) \; (for \; k \gt 2)\\ &= 6(k+1)^2\end{aligned} \end{equation}\]

Hence  
\[P(k+1)\]
  is true and the inequality is proved.