Find the smallest $N \in Z^+$ satisfying the property that $n^2 \leq 2^n$ for all $n \geq N$, and prove your result using induction.
I assume this can be done simply by testing it will random N values until one is found that doesn't satisfy the condition that $n \geq N$, but is there a more efficient mathematical way to solve this?
We see than $n=4$ works, and that $n=3$ doesn't. We can see that for $n>1$, $2n^{2}>(n+1)^{2}$ (this can be seen by expanding, and rewriting as $(n-1)^{2}>2$, which is true for all $n>2$). Now, as we have that, the induction is trivial: If for some $n$, $n^{2}\leq 2^{n}$ (and $n>2$), $2n^{2}<2^{n+1}$, but since $(n+1)^{2}<2n^{2}$, we prove by induction that $4$ is our answer.