Prove that $n! < n^n $ where n >1 and is an integer. Lets skip the base case cause its trivial.
Assume that: $$ k! < k^k = $$ Inductive step: $$(k+1)! < (k+1)^{k+1} =$$ $$(k)!(k+1) < (k+1)^k (k+1) =$$ $$k! < (k+1)^k$$
why do some of my classmates say the way i solved it is wrong? am i really wrong?
On
Your proof is basically correct, just lacks words. I took your proof from above and just added a few words here and there between [...], and a bit extra towards the end:
Lets skip the base case cause its trivial. [Fair enough, though in real life do make sure to include the proof]
[Inductive step: (this is where the inductive step begins, not below) ]Assume [for $k> 1$] that: $$ k! < k^k $$ [We need to show that:] $$(k+1)! < (k+1)^{k+1} $$ [which amounts to ] $$(k)!(k+1) < (k+1)^k (k+1) $$ [which is the same as]
$$k! < (k+1)^k$$
[which holds since $k!<k^k<(k+1)^k$].
Unfortunately you have made several mistakes in attempting to perform a proof by induction.
Firstly, you have not proven the base case ($n=2$). This may seem trivial, but it still needs to be shown.
Secondly, your inductive section begins with $(k+1)!<(k+1)^{k+1}$ but we don't know this- it's what we have to prove! You can only begin with things you know are inherently true or more commonly, what you have assumed to be true. Then from this you need to somehow logically make your way to what you want to show.
Notice also that you never actually seem to use your assumption in the proof and your last line seems to just hang there without any real explanation of its significance.
You probably have the majority of the ingredients there to pull of this proof, but you need to be much more careful!