Prove the $3^n < n!$ for all $n > 6$ I'm trying to use induction to prove this.  I'm sure it's a simple proof, but I can't seem to get over the first few steps.  Any help?
Allow $P(n)=3^n<n!$
Base Case:
$P(7) = 3^7<7! \rightarrow$ True.
Induction:
Assume $P(k) = 3^k<k!$
Now we must prove $P(k+1)$.  Here's where I'm lost.  If I'm adding a +1 to the exponent on the LHS, where would I add it to the factorial on the RHS?
 A: The key to induction proofs is finding a way to work your induction hypothesis into the "$k+1$" case.
We want to show $3^{k+1} < (k+1)!$. Since you know $3^k < k!$, we need to keep an eye out for a factor of $3^k$. Let's just start with the lefthand side of the "$k+1$" case and see what we can do.
$$
\begin{align*}
3^{k+1} &= 3 \cdot 3^k\\
&< 3 \cdot k! && \text{(inductive hypothesis)}\\
&< (k+1) \cdot k! && \text{(since k > 2)}\\
&= (k+1)!
\end{align*}
$$
A: I'm not sure I understand your question. The rest of the comments/answers have interpreted it in one way, but my answer to the question:

Here's where I'm lost. If I'm adding a +1 to the exponent on the LHS, where would I add it to the factorial on the RHS?

Is that the RHS will look like $(k+1)!$. I'm not sure why you might have thought otherwise? Did you think it might be $k! + 1$? Remember that you are inducting on $k$. So wherever you saw a $n$ you'd replace it with $k+1$.
If this was indeed the confusion you had it would be really really worth it to make sure you understand induction better (and maybe even refresh yourself on the definition of a function?). I can write up a loose description of induction that might help you if you want it.
A: Single line:
If $3^k < k!$ and $k>6$ then $3^{k+1} = 3\times 3^k < 3\times k!  < (k+1)\times k! = (k+1)!$
(As $n > 6$ we can assume than $k>6$ and therefor $k+1 > 7 > 3$)
=======================================

If I'm adding a +1 to the exponent on the LHS, where would I add it to the factorial on the RHS?

What does "adding a $+1$ to an exponent" mean?
It means multiplying by the base.
So do that.
$3^{k+1} = 3\times 3^k$.
Now if $3^k < k!$ that means $3\times 3^k < 3\times k!$

where would I add it to the factorial on the RHS?

well that means multiplying the factorial by $k+1$.
So $(k+1)! = k! \times (k+1)$.
Now you have to compare $3\times k!$ to $k! \times (k+1)$ and the rest is obvious.
.....
$3\times k! < (k+1)\times k!$ so long as $k + 1 > 3$ or $k > 2$.  And as $k > 6$ we have that.
.....
So
$3^{k+1}\color{tan}=$

$\color{red}{3\times 3^k<3\times k! < (k+1)\times k!}$

$\color{tan}= (k+1)!$
