Is Induction Proof for 2^n>n Legitimate?

[altcmdesc]

I'm just wondering if this is a legitimate proof by induction.

Prove, for all natural numbers n, that 2^n>n

Proof. For n=0, 2^0=1>0 and for n=1, 2^1=2>1. Similarly, if n=2, then 2^2=4>2. Now assume n>2 and we have proven the result for n-1. We must show it is true for n.

We have:

<br /> 2^{n-1}&gt;n-1<br />

<br /> 2^n&gt;2(n-1)=(n-1)+(n-1)&gt;(n-1)+1=n<br />

The last inequality follows from the fact that n&gt;2.

I'm worried about the fact that I have more than one base case. Is this still alright?

 

[mutton]

Looks right. Doesn't matter how many base cases you have as long as you cover all natural numbers in the end.