Order of growth proofs? I was wondering how people go about showing the proofs with orders of growth? Currently, I have the following functions and I know what order they go in, but I'm not sure how to prove them. I simply put them into a graphing application I saw online and the graph itself obviously showed the growth.
$$\begin{align}
& 2^{\sqrt{\log n}}\\
& 2^n\\
& n^{4/3}\\
& n  \log^3 n\\
& n^{ \log n}\\
& 2^{2^n}\\
& 2^{n^2}
\end{align}$$
For example, I know that $2^{2^n}$ has the largest growth rate, with $2^{n^2}$ coming in second. I know this because if I plug the value of $5$ it turns into $2^{32}$ vs $2^{25}$, $6$ is $2^{64}$ vs $2^{36}$. But, is that the proof itself?
As far as the odd ones out, I have no idea how I could show a proof comparison of $n^{4/3}$ vs $n\log^3 $n vs $2^n$, etc.
 A: General Rules of Thumb which hold for sufficiently large $n$:
$$\begin{align}
&\log^c(n) \leq n^{\epsilon}, \quad \text{ for any } \epsilon > 0, \quad \text { and any constant } c
\\

\\
&n^k \leq k^n, \qquad \text{where } k \text{ is any constant } k > 1
\end{align}
$$
As Chris noted above, we can write this as
$$
A \ll \log^c (n) \ll n^{\epsilon} \ll k^n
$$
To actually prove these rules, you just compute limits (assuming you know calculus).  To see that $\log (n) \ll n^{\varepsilon}$ for example, just note that
$$
\lim_{n \to \infty} \frac{\log n}{n^{\epsilon}} = \lim_{n \to \infty} \frac{n^{-1}}{\epsilon n^{\epsilon - 1}} = 0 = \lim_{n \to \infty} \frac{1}{\epsilon n^{\epsilon}} = 0.
$$
where $A$ and $c$ are constants, $\epsilon > 0$, and $k$ is a real number such that $k > 1$.
If you want to show that $\log^c(n) \ll n^{\epsilon}$ for any positive integer $c$, then just note that
$$
\lim_{n \to \infty} \frac{\log^c (n)}{n^\epsilon} \leq \left( \lim_{n \to \infty} \frac{\log (n)}{n^{\epsilon/c}} \right) \dots \left( \lim_{n \to \infty}  \frac{\log(n)}{n^{\epsilon/ c}}\right) = 0
$$
by what we showed above.  We can actually take $c$ be to be any arbitrary real number and the above still holds.  This is easy to see since every real number is less than or equal to some integer (infinitely many in fact!).  I will leave it to you to show that $n \log^3 (n) \ll n^{4/3}$, for example.  The more complicated functions work the same way but we might need to be slightly creative when evaluating the limits.
Hope this helps.
