Prove by induction: $(n-1)!>2^n$ for all $n \geq 6$ Let $P(n)$ be the statement
$(n-1)!>2^n$
Use mathematical induction to show that $P(n)$ is true for all $n \geq 6$.
I've got $(k)!>2(k-1)!$. With the given value of $k= 6$ is this proof enough?
 A: Induction needs a base case as well as an induction step.
The single fact you stated is enough for most of us to fill in the rest of the details of the inductive proof. But since this appears to be an exercise for your studies, in which you are (presumably) expected to demonstrate that you understand how mathematical induction works, you should fill in all the details yourself, and not assume that the reader will guess how you would have finished the proof.
You should therefore show the base case explicitly, and you should show step by step how to prove that $P(k+1)$ is true if $P(k)$ is true and $k\geq 6.$
What you showed is just one sub-step of the inductive part of the proof.
A: Yes, since $k!=k(k-1)!>2(k-1)!$ for $k>6$.
A: As a proper proof by mathematical induction, you need to have an explicit base and step, and show a little more work and explanation. I mean, if you were to hand this in as your answer on your HW, and if I were your instructor, I would take points off. So:
Base: $n = 6$.  
Then $(6-1)! = 5! = 120$  And $2^6 = 64$. And since $120 >64$, we indeed have that $(n-1)!>2^n$ for n=6
Step: Take some number $n \ge 6$ and suppose (inductive hypothesis) $(n-1)!>2^n$
Now we have to show that: $((n+1)-1)!>2^{n+1}$
Let's see:
$((n+1)-1)! = n! = n(n-1)! >$  (since $n \ge 6$)
$2(n-1)! > $ (inductive hypothesis)
$2*2^n = 2^{n+1}$
Also good!
OK, we got our base, and we got our step, so the result is proven by induction.
A: David K's answer makes a good point, especially about the base case for your problem. In particular, if we simply jump straight to the inductive part of the proof, without bothering with the bast case, then it would seem that
$$
k! =k(k-1)!>k\cdot 2^k > 2\cdot 2^k=2^{k+1}
$$
only requires $k\geq3$. But note that the first time your statement holds is when $n=6$.
