How can I prove that $2^n\ge n^4$?

Question

I'm trying to prove the following inequality. $$2^n\ge n^4 (n>n_0)$$ I noticed that from $n = 20$, the following inequality holds, and I want to use induction for the proof. However I simply cannot find any way to prove whenever $n$ holds, $n+1$ also holds as well. Any help would be appreciated. Thanks!

Answer 1

You can use induction to show that $2^n \geq n^4$ for all $n \geq 16$. When $n = 16$, we have $2^{16} = 16^4$. Suppose that $2^n \geq n^4$ for $n \geq 16$. We will show that $2^{n+1} \geq (n+1)^4$: $$ (n+1)^4 = \left(1 + \frac{1}{n}\right)^4 \cdot n^4 \leq \left(1 + \frac{1}{16}\right)^4 \cdot 2^n < 2 \cdot 2^n = 2^{n+1}. $$

Answer 2

$$2^n>n^4\iff 2^{n/4}>n.$$

Then

$$2^{n/4}>n\implies 2^{(n+1)/4}>\sqrt[4]2\,n>n+1$$ if $$n>\frac1{\sqrt[4]2-1}>5.$$

Answer 3

Because $$2^n=(1+1)^n>\binom{n}{9}>n^4$$ for all $n\geq20$.

Also, $$2^n>\binom{n}{5}>n^4$$ for all $n\geq130$.