Proving $n! > n$ for $n > 2$ using mathematical induction

Question

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!

Answer 1

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...

Answer 2

Since in your case $K > 2$, the statement $K!+1 < (K+1)!$ is true. This is because $(K+1)! = K!(K+1) = K!K + K!$, and $K!K > 1$ for $K > 1$.

Answer 3

Starting with $K \lt K!$,

multiplying both sides by $K+1 \gt 0$ you have $K(K+1) \lt (K+1)!$,

but since $K\gt 2$ you have $K+1 \lt K(K+1) $ and so $K+1 \lt (K+1)!$