Question: Consider the product
$$\prod\_p\Big(1+\frac1{(p-1)^3}\Big)$$
over the primes. I can't easily see why this should converge. Something like $\prod\_p\big(1+\frac 1{p-1}\big)$ doesn't converge, since (I believe) $\prod\_{p\leqslant x}\big(1+\frac 1{p-1}\big)\sim\log\log x$.

I've rewritten the product as
$$\exp\bigg(\sum\_{p}\sum\_{n}\frac{(-1)^{n+1}}{n(p-1)^{3n}}\bigg)$$
using the Taylor series for log, but I'm not sure how to continue.

Answer: 
Let $p$ denote the $n$th prime so$$p-1\ge n\implies1+(p-1)^{-3}\le\exp n^{-3},$$so your product $\le\exp\zeta(3)$.
Since,
for $x > 0$,
$\ln(1+x)
=\int\_0^x \dfrac{dt}{1+t}
$,
$\ln(1+x)
\lt x$
and
$\ln(1+x)
\gt \dfrac{x}{1+x}
\gt x-x^2
$.

Therefore

$\begin{array}\\
\prod\_p\Big(1+\frac1{(p-1)^3}\Big)
&=e^{\sum\_p \ln\Big(1+\frac1{(p-1)^3}\Big)}\\
&\lt e^{\sum\_p \frac1{(p-1)^3}}\\
\end{array}
$

and since
$\sum\_p \frac1{(p-1)^3}$
converges,
$e^{\sum\_p \frac1{(p-1)^3}}
$ converges.

More generally,
if
$a\_n > 0$ and
$\sum a\_n$ converges then

$\begin{array}\\
\prod\_n\Big(1+a\_n\Big)
&=e^{\sum\_n \ln(1+a\_n)}\\
&\lt e^{\sum\_n a\_n}\\
\end{array}
$

so
$\prod\_n\Big(1+a\_n\Big)$
converges.