Can't prove $2^n > n$ with Mathematical Induction As the title states, I have a problem with proving $2^n > n$
I can do the basis step fine:
Basis step:
"n = 1"

2^1 = 2
2 > 1

So it is true for $n$.
But then I get stuck with the inductive step. Here's how I tried, but then I get stuck. I don't know what to do. 
Inductive step:
"Assume true for n"
"Prove for n+1"

  2^(n+1) > n+1

= 2^1 * 2^n
= 2*2^n
= 1*2^n + 2^n
= 2^n + 2^n > n+1

If you know the answer, please explain it to me so I know how to do it in the future. Any help is appreciated.

For $n=1$, we have $2^1=2>n=1$.  Now, assume that for some $k>1$ we have 
$$2^k>k \tag 1$$
Then, we have 
$$\begin{align}
2^{k+1}&=2(2^k) \tag 2\\\\
&>2(k) \tag 3\\\\
&=k+k \tag 4\\\\
&>k+1 \tag 5
\end{align}$$
for $k>1$.  

Notes:
In going from $(2)$ to $(3)$, we used the assumption $2^k>k$ as given in $(1)$.  
Then, in arriving at $(4)$, we simply used the fact that $2k=k+k$.  

Finally, since $k>1$, $k+k>k+1$.
Therefore, by the principal of mathematical induction, $2^n>n$ for $n\ge 1$.
