Prove by induction that $n^3 < 3^n$ The question is prove by induction that $n^3 < 3^n$ for all $n\ge4$. 
The way I have been presented a solution is to consider:
$$\frac{(d+1)^3}{d^3} = (1 + \frac{1}{d})^3 \ge (1.25)^3 = (\frac{5}{4})^3 = \frac{125}{64} <2 < 3$$
Then using this $$(d+1)^3 = d^3 \times \frac{(d+1)^3}{d^3} < 3d^3 < 3 \times 3^d = 3^{d+1}$$ so we have shown the inductive step and hence skipping all the easy parts the above statement is true by induction. 
However I don't find this method very intuitive or natural; is there another way to attack this problem? 
The approach I wish to take involves starting from $$ 3^{d+1} = 3 \times3^d > 3d^3$$ but then I do not know how show further that $3d^3 > (d+1)^3 $ to complete the inductive step. I have looked around at the proofs related to showing that $2^n > n^2 $ inductively for $n \ge 5$ but cannot relate the proof for that case to this case.
Also, is there a more general method that could be used to solve, say $a^n > n^a $ for $ n \ge k $ for some $k\in \Bbb R$

This base case holds because $4^3 < 3^4$. 
To show that the inductive step holds, we need to show that: 
$(n + 1)^3 < 3^{n + 1}$ holds if $n^3 < 3^n$ holds. 
Note that: 
$3^{n + 1} = 3 * 3^n > 3n^3$(since $3^n > n^3$ by the inductive hypothesis) > $(n + 1)^3$. 

By binomial expansion: $(n + 1)^3 = n^3 + 3n^2 + 3n + 1$. 
So: 
$3n^3 ≥ (n + 1)^3\Leftrightarrow 3n^3 ≥ n^3 + 3n^2 + 3n + 1 \Leftrightarrow 2n^3 - 3n^2 - 3n - 1 ≥ 0. $
Since $2n^3 - 3n^2 - 3n - 1 = 0$ has a real solution at about $n \approx 2.26$ and $f(3) > 0$, we see that $2n^3 - 3n^2 - 3n - 1 > 0$ holds on the interval $(2.26,\infty)$. Then, because $4^3 < 3^4$, we see that: 
$$3^{n + 1} > (n + 1)^3$$ for all $n \geq 4$, which is what we wanted to show. 

Hope it helps.
