= Solution
For $x\geq2$, expand the finite <Euler product> as a product of <geometric series>:
$$
P(x):=\prod_{p\leq x}\left(1-\frac1p\right)^{-1}
=\sum_{\substack{n\geq1\\p\mid n\Rightarrow p\leq x}}\frac1n.
$$
Every integer $1\leq n\leq x$ has all its prime factors at most $x$, so
$$
P(x)\geq\sum_{n\leq x}\frac1n.
$$
The <harmonic series> diverges, and therefore
$$
\boxed{\prod_p\left(1-\frac1p\right)^{-1}=\infty.}
$$
Suppose instead that $\sum_p1/p$ converged. For $0<t\leq1/2$,
$$
0\leq-\log(1-t)-t=\sum_{j\geq2}\frac{t^j}{j}\leq\frac{t^2}{1-t}\leq2t^2.
$$
Since $\sum_p1/p^2$ converges, this would make
$$
\log P(x)=\sum_{p\leq x}-\log\left(1-\frac1p\right)
$$
converge to a finite limit, contradicting the divergence of $P(x)$. Hence the <Euler proof that the sum of reciprocals of primes diverges> gives
$$
\boxed{\sum_p\frac1p=\infty.}
$$
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