Prove by induction that for all $n \geq 1$.

First, prove this proposition is true for the base case $n=1$.

True for base case, $n=1$. Clearly $2>1$.

Next, assume true for some arbitrary number $k$. Assume true for $n=k$ :

is true for $k \geq 1$.

Finally, prove for $n=k+1$.

Use the definition for the powers of $2$.

By the inductive hypothesis, $2^k > k^2$, so:

Now compare $2 \cdot k^2$ and $(k + 1)^2$:

We need to show :

which simplifies to :

By mathematical induction, $2^n > n^2$ holds for all $n \geq 1$.