Proving 2^n > n^3 for all n > 9 (updated)

by Nibir Sankar, Jun 26, 2025

Prove $2^{n} > n^{3}$ for all $n > 9$

Proof:

By mathematical induction.

Let $A (n)$ denote $2^{n} > n^{3}$ .

For $n = 10$ (base step)

\begin{array}{r} 2^{10} > 10^{3} \\ 1024 > 1000 \end{array}

Thus, $A (10)$ is true.

Now, assuming $A (n)$ is true, we get:

\begin{array}{r} 2^{n} > n^{3} \end{array}

Multiplying by $2$ on both sides:

\begin{aligned} 2^{n} \cdot 2 > 2 n^{3} \\ ⟹ & 2^{n + 1} > 2 n^{3} (i) \end{aligned}

We want to deduce, for $A (n + 1)$ :

\begin{array}{r} 2^{n + 1} > (n + 1)^{3} \end{array}

To prove this, we have to prove that: (from $(i)$ )

\begin{array}{r} 2 n^{3} \geq (n + 1)^{3} (ii) \end{array}

( $\forall n > 9$ )

We do this by contradiction.

Assuming $(ii)$ is false, then $2 n^{3} < (n + 1)^{3}$ must be true for all $n > 9$ .

But for $n = 10$ ,

\begin{array}{r} 2 (10)^{3} = 2000 > (10 + 1)^{3} = 1331 \end{array}

Contradiction.

Thus, $(ii)$ is true, and which follows that $(i)$ must be true.

Consequently:

\begin{array}{r} 2^{n + 1} > (n + 1)^{3} \end{array}

Which follows that $A (n)$ must be true.

Therefore, by the principle of mathematical induction, we have proven that $2^{n} > n^{3}$ for all $n > 9$ .