Prove by mathematical induction that $\frac{n^3}{n!} < \frac{1}{2^n}$ for some $n\geq m$ where $m > 0$
To start off, I assume the hypothesis that it is true for $n$.
Trying to prove for $n+1$,
$$\frac{1}{2^{n+1}} - \frac{(n+1)^3}{(n+1)!} > 0.$$
Therefore,
$$\frac{1}{2^{n+1}} - \frac{(n+1)^3}{(n+1)!} = \frac{1}{2^n} \times \frac{1}{2} - \frac{(n+1)^3}{(n+1)!} > \frac{1}{2} \times \frac{n^3}{n!} - \frac{(n+1)^3}{(n+1)!} = \frac{1}{2n!} [ n^3 - 2n^2 - 4n - 2]$$
So basically, my task now is to prove that the last term is greater than $0$. Am i on the right track and how do I go further?
Proof by indution:
As the base case, let's take $m=10$: $$ \frac{10^3}{10!}-\frac{1}{2^{10}}\approx -7.0\times 10^{-4}\implies \frac{10^3}{10!}<\frac{1}{2^{10}}. $$ Now, supposing the inequality is true for $n\geq 10$, let's prove that it is also true for $n+1$. Indeed, \begin{align} \frac{(n+1)^3}{(n+1)!}&=\frac{(n+1)^2}{n^3}\frac{n^3}{n!} &\phantom{}\\ &<\left(\frac{1}{n}+\frac{2}{n^2}+\frac{1}{n^3}\right)\frac{1}{2^n}&[\text{by the induction hypothesis}] \\ &\leq\left(\frac{1}{10}+\frac{2}{10^2}+\frac{1}{10^3}\right)\frac{1}{2^n}&[\text{because $n\geq 10$}] \\ &<\frac{1}{2^{n+1}}. \quad{\square} &[\text{because $(\ldots)<1/2$}] \end{align}