Proof that $n^3 \leq 2^n$ for $n \geq 10$ by induction 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?

Your solution looks good, but you might want to add more details as to why
$$(k+1)^3\leq k^3+k^3$$
This can be seen explicitly as
$$(k+1)^3=1 + 3 k + 3 k^2 + k^3=k^3+k^3\left(\frac{3}{k}+\frac{3}{k^2}+\frac{1}{k^3}\right)$$
Since $k\geq 4$ we have
$$k^3+k^3\left(\frac{3}{k}+\frac{3}{k^2}+\frac{1}{k^3}\right)\leq k^3+k^3\left(\frac{3}{4}+\frac{3}{4^2}+\frac{1}{4^3}\right)=k^3+\frac{61}{64}k^3<k^3+k^3$$
