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?

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.
