Hey guys so I think I have completed this proof but I'm not sure if its valid. Here it is:
Prove that $ 2^n < (n+1)! \quad\text{for}\quad n >= 2 $
Here is my proof:
Base Case P(2) = $ 4 < 6 $
Inductive Hypothesis (IH) P(k) = $ 2^k < (k+1)! $
Proof P(k+1) = $ 2(k+1) < (k+2)! $
$ 2 * 2^k < (k+1)! * (k+2) $
I have already shown that $ 2^k < (k+1)! $ is true by IH. With $2$ being multiplied on the left and $(k+2)$ being multiplied on the right, if I can prove that $2 < (k+2) $ than the whole equation is true.
2 is always less than $k+2$ because k must be greater than or equal to 2 so the equation at minimum is $2 < 4$ .
End Proof
Is this valid? And if not, what am I missing? why is this approach wrong?
Several of the important pieces of the inductive proof are there, but everything could be written up much more smoothly. For example, your inductive hypothesis should be to "fix some $k\geq2$ and assume that $P(k) : 2^k<(k+1)!$ holds." This addresses Andre's comment--he "did not like" $P(k)=2^k<(k+1)!$ because that is technically incorrect. Why? Well, $P(k)$ is supposed to denote a proposition or a statement, not an equivalence.
That being said, I think you could more effectively write up the main part of your induction proof as follows: \begin{align} 2^{k+1}&=2\cdot2^k\tag{law of exponents}\\[1em] &< 2\cdot(k+1)!\tag{by induction hypothesis}\\[1em] &< (k+2)\cdot(k+1)!\tag{since $k\geq2$}\\[1em] &= (k+2)!\tag{by definition} \end{align}