How can I prove $2^n > n^2 $ by induction using a basis $> 4$ I've been trying to prove this statement by induction; however, in following the steps I normally take I end up utterly stuck. I know that I must be missing something, but I have been stuck on this for a while now and can't seem to figure it out. Does anyone know how I can get to a sum on the RHS of my proof statements? Any other advice on how to find this?
Prove $**2^n > n^2**$ by induction using a basis > 4:
Basis: $**n = 5** **2^n > n^2**$
                 32 > 25
Assume: $**2^n > n^2**$
Prove:  $**2^(n+1) > (n+1) ^2**$
$**2^{n+1} = 2^n * 2** $    LHS Prove = LHS Assume + Fix 
$**2^{n+1} > (n^2) *2**  $   Utilize assumption
$**2^{n+1} > 2(n^2)** $      Could utilize assumption again... 
$**2^{n+1} > 1(n^2)**  $       ..doesn't seem to help
$**2^{n+1} > (n * 1)^2**   $     Basically just$ n^2$ 
 A: For sure this question has been asked so many times  here...
As this is your first question it would be kind of rude if i vote to close this..
Coming to the Question :
You have $2^n>n^2$
You have to prove $2^{n+1}>(n+1)^2$
So, you could consider (as you did)
$2^{n+1}=2^n.2>2n^2$
Suppose you prove $2n^2>(n+1)^2$ then you are done..
I guess you know  what is $(n+1)^2$!!
Compare $(n+1)^2$ (just by expanding it) with $2n^2$
Then you are done!
A: You are on the right track. Under the inductive hypothesis, you've shown that $2^{n+1} > 2n^2$.
If you could show that $2n^2 \ge (n+1)^2$, you would be done.
This is equivalent to $n^2 - 2n - 1 \ge 0$, or $(n-1)^2 \ge 1$. That's definitely true!
A: You can also show this directly without using induction.
Define $f:(0,\infty)\rightarrow (0,\infty)$ by $f(x)=x^{\frac1x}$. Then it is easy to compute $f'(x)=x^{\frac1x-2}(1-\log x)$, so that $f$ is strictly decreasing for $x>e$. This means that $5^{\frac15}>n^{\frac1n}$ for $n=6,7,8,\ldots$; and since one can easily compute $2^{\frac12}\approx 1.41 >1.38\approx 5^{\frac15}$, we have (for integral $n\geq 5$)
$$2^{\frac12}>n^{{\frac1n}}$$
$$\left(2^{\frac12}\right)^{2n}>\left(n^{{\frac1n}}\right)^{2n}$$
$$\boxed{2^n>n^2}$$
for integral $n\geq 5$ as desired.
