TITLE: Proving $n! > n$ for $n > 2$ using mathematical induction
QUESTION [5 upvotes]: I have to prove $n<n!$ for all $n>2$ by mathematical induction. 
I did it as follows. I proved the base case. 
Then let it be true for $K>2$:
$$ K<K! $$
I have to prove,
$$ K+1<(K+1)! $$
$$ K<K! $$
Adding $1$ on both sides
$$ K+1<K!+1<(K+1)! $$
Hence
$$    K+1<(K+1)! $$
Is the last step I did ($K!+1<(K+1)!$) Right? 
Please help me out. Thanks!

REPLY [8 votes]: Why use induction at all when you can see plainly that $n! = n\cdot (n-1) \cdots 1 \ge n\cdot (n-1) > n$ if $n>2$ ?
If this is an exercise, it's a silly one. It's exercises like this that give a bad name to induction...