If $f(1) = 2$ and $f(n) = n \cdot f(n-1)$ then $f(n) \gt 2^n$ for all $n \gt 2$ I'm having a little difficulty in proving what are probably simple induction proofs. Here is the question.
Define function $f(n)$ as follows. $f(1) = 2$ and $f(n) = n\cdot f(n-1)$ when $n > 1$.
Use induction to prove that $f(n) > 2^n$ for all $n>2$. 
Note that $f(1) =2$, $f(2)=4$. I understand the basis step, so I'm not going to write that out.
Now we have. $f(k) = k \cdot f(k-1)> 2^k$ we want to show the following.
$f(k) + f(k+1) = (k+1)\cdot f(k)> 2^{k+1}$ This is where I'm getting stuck. Can someone give me a hint as to how to proceed?

$$\begin{align}
P(3):&\quad f(3)=12 > 8 = 2^3 \\
P(k):&\quad f(k)>2^k \\
P(k+1):&\quad f(k+1) = (k+1)f(k) > (k+1) 2^k >2 \cdot 2^k = 2^{k+1}
\end{align}$$
So what we do is first to show that $f(n)>2^n$ for $n=3$. Then we assume that it is also true for $n=k$. At last we prove that it is true for $k+1$ if it is true for $k$. 
What I did in the last step is this:


*

*I wrote down the expression for $f(k+1)$.

*Because of the step $P(k)$ we know (assume) that $f(k)>2^k$ which implies that $(k+1)f(k) > (k+1) 2^k$ since $k+1>0$.

*We also know that $k+1 > 2$ which implies that $(k+1) 2^k >2 \cdot 2^k$.

*$2 \cdot 2^k = 2^{k+1}$. So it follows that $f(k+1)>2^{k+1}$.

