discrete mathematics - How to prove that $4^n>n^2$ using induction...

today I've encountered a question like the following;
$$\text{Prove that }4^n>n^2\text{ using induction.}$$
My Attempts:

I have realised that this works for $P(1)$, my next attempt was $p(n)\implies p(n+1)$....(1)

I have tried to multiply both sides with a $4$ which gave $4^{n+1}>4n^2$ I have tried to turn it out like $4>1^2$ and that gave me $4^{n+1}>n^2\cdot1^2$.....(2)

After that pointless attempt I've added $2n+1$ to both sides but I couldn't figure out still what $2n$ goes to in the left side...(3)

What are your suggestions?

With the real question being the first one, is there any other way to prove this numerically? (Perhaps in a more entertaining way?:))

Answer

Numerically is not a proof.

Induction works in this way

1. It is true for $n=1$


2. Suppose that is is true for $n>1$, prove it for $n+1$


3. It is true for any $n\in\mathbb{N}$.

proof

1. actually $4^1> 1^2$


2. (I.H.) if $4^n>n^2$ for $n>1$ consider that $4^{n+1}=4\cdot 4^n$. Now use the Inductive Hypothesis (I.H.)

$4\cdot 4^n > 4\cdot n^2 =2^2 \cdot n^2=(2n)^2>(n+1)^2$ as $n>1$

proved, so

3. For any $n\in \mathbb{N}$ we have $4^n>n^2$

QED
$$
.
$$

To say the truth $4^n > n^{1000}$ for $n>6312$

Indeed induction can start from any $n$, but this is another story