On Wikipedia at link currently is:
\begin{align}
\ln \left( \sum_{n=1}^\infty \frac{1}{n}\right) & {} = \ln\left( \prod_p \frac{1}{1-p^{-1}}\right)
= -\sum_p \ln \left( 1-\frac{1}{p}\right) \\
& {} = \sum_p \left( \frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \cdots \right) \\
& {} = \left( \sum_{p}\frac{1}{p} \right) + \sum_p \frac{1}{p^2} \left( \frac{1}{2} + \frac{1}{3p} + \frac{1}{4p^2} + \cdots \right) \\
& {} < \left( \sum_p \frac{1}{p} \right) + \sum_p \frac{1}{p^2} \left( 1 + \frac{1}{p} + \frac{1}{p^2} + \cdots \right) \\
& {} = \left( \sum_p \frac{1}{p} \right) + \left( \sum_p \frac{1}{p(p-1)} \right) \\
& {} = \left( \sum_p \frac{1}{p} \right) + C
\end{align}
and since $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges, so must $\sum_{p} \frac{1}{p}.$
Currently there is language like "audacious leaps of logic" and "correct result by questionable means".
Wouldn't this be a valid proof if we reversed it, though? If we assume $\sum_p \frac{1}{p}$ converges, can't we just go through the steps backwards and find $\sum_{n=1}^{\infty} \frac{1}{n}$ converges, contradiction? Euler's work seems reasonable to me.
Best Answer
$ \ln \left( \sum_{n=1}^\infty \frac{1}{n}\right)$ has no meaning.
It is like you say $$\ln (\infty)=...$$
But $\ln x$ is defined on real numbers and not on "anything we understand".
But everyone understands that Euler "understood" that the method was correct.
This is why the proof is attributed to him.