I have this statement:
Which I represent using the predicate P(n).
In order to prove this I have to provide a 'first' case which satisfies this expression. Which in this example is 4. Because every value greater than or equal to 4 will indeed support the statement. Then prove the fact for n+1... but I am really confused about how is the proof structure.
I need to write a proof... and in proofs the steps have to be justified, and choosing a random value of n = 4 won't be ok without its respective justification, so what could I put there?
When $n=4$, we have $2^n=2^4=16$ and $n!=4!=24$, so the base case holds. Now, assume $2^k<k!$ for some $k\ge4$.
We wish to show that $2^{k+1}<(k+1)!$. So, $$2^{k+1}=2*2^k<2*k!<(k+1)*k!=(k+1)!$$
This follows from the assumption that $2^k<k!$ and the fact that $2<k+1$, since $k>4$.
Thus, by induction, $2^n\lt n!$ for all $n\ge 4$.
Now, show that this is not true when $n=3$. Thus, we have a strict base case.