If $n \ge 1$, does it follow that $\left(\frac{3}{2}\right)^n > n$ It seems to me the answer is yes.  Here's my reasoning:
For $n=1$, $\frac{3}{2} > 1$
and, $\frac{d}{dn}\left(\frac{\left(\frac{3}{2}\right)^n}{n}\right) > 0  $
Using the quotient rule:
$\dfrac{d}{dn}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} = \frac{\left(\frac{3}{2}\right)^n\left(\ln \frac{3}{2}\right)n - \left(\frac{3}{2}\right)^n }{n^2} > 0$ for $n \ge 1$
Am I correct?  Is there a more standard way to show this?
 A: We can observe that
$$\left(\frac{3}{2}\right)^n > n\iff 3^n >n2^n$$
and then proceed by induction.
A: Here is an approach that is similar to that in your question, but perhaps a bit simpler.

Since
$$
\frac{\mathrm{d}}{\mathrm{d}x}\frac{\log(x)}x=\frac{1-\log(x)}{x^2}\tag1
$$
The function $\frac{\log(x)}x$ is increasing for $x\lt e$ and decreasing for $x\gt e$ and thus has a maximum of $\frac1e$.
The inequality in the question is equivalent to
$$
\frac{\log(n)}n\lt\log\left(\frac32\right)\tag2
$$
which is true since $\frac1e\lt\log\left(\frac32\right)$.

Verification that $\boldsymbol{\frac1e\lt\log\left(\frac32\right)}$
If we can assume that $\frac52\lt e\lt\frac{11}4$, then
$$
\frac1e\lt\frac25\tag3
$$
Furthermore,
$$
\begin{align}
e^2
&\lt\left(\frac{11}4\right)^2\\
&=\frac{121}{16}\\
&\lt\frac{243}{32}\\
&=\left(\frac32\right)^5\tag4
\end{align}
$$
Combining $(3)$ and $(4)$ shows that
$$
\frac1e\lt\frac25\lt\log\left(\frac32\right)\tag5
$$
A: Just check $f(x) = (\frac{3}{2})^x - x > 0$ for $x=1,2,3$.
Then for $x> 3$ you have $f'(x)= (\frac{3}{2})^x\ln \frac{3}{2} -1> 0$ anyway.
