TITLE: Proof that $n^3 \leq 2^n$ for $n \geq 10$ by induction
QUESTION [6 upvotes]: I'm trying to prove this statement and it got a bit clumsy but at least it seems to make sense. I'd like to know if this proof is valid. Here's what I managed to do:
First the base case:
$n = 10$
$$n^3 \leq 2^n$$
$$1000 \leq 1024$$
I assumed that it's true for $k$:
$$k^3 \leq 2^k$$
Then I tried to prove that it's true for $k+1$. First I multiplied both sides by $2$:
$$2k^3 \leq 2^k \cdot 2$$
$$2k^3 \leq 2^{k+1}$$
$$2k^3 \leq 2^k + 2^k$$
$$k^3 + k^3 \leq 2^k + 2^k$$
Since $n \geq 10$:
$$(k+1)^3 \leq k^3 + k^3 \leq 2^k + 2^k$$
therefore $$(k+1)^3 \leq 2^{k+1}$$
Can someone confirm this, please?

REPLY [4 votes]: Just to give one more proof of the crucial missing argument,
$$(k+1)^3=k^3+3k^2+3k+1\le k^3+3k^2+3k^2+k^2=k^3+7k^2\le k^3+k^3$$
The first inequality is satisfied for all $k\ge1$, and the second for all $k\ge7$.