Why is Euler's statement $\exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots$ true? Euler, in his paper Variae observationes circa series infinitas [src], makes the following statements in his Theorem 19.
$$ \exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \frac{1}{4}D + \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \dots $$
Where he defines:
$$A = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots $$
$$B = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots $$
$$C = \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots $$
Question: I can't see how that statement is true.
I would appreciate replies that don't assume university level training in mathematics.
 A: It can be shown that
\begin{align*}
\sum\limits_{n = 1}^\infty  {\frac{1}{{n^s }}} & = \prod\limits_p {\frac{1}{{1 - \frac{1}{{p^s }}}}}  = \exp \left( { - \sum\limits_p {\log \left( {1 - \frac{1}{{p^s }}} \right)} } \right) \\ &= \exp \left( {\sum\limits_p {\sum\limits_{n = 1}^\infty  {\frac{1}{{np^{ns} }}} } } \right)=
\exp \left( {\sum\limits_{n = 1}^\infty \frac{1}{n} {\sum\limits_p {\frac{1}{{p^{ns} }}} } } \right),
\end{align*}
whenever $\Re s>1$. Taking the limit $s\to 1+$ yields Euler's formal identity. I say formal, because the expressions on both sides will diverge. Note that I am not making an attempt to describe this in terms of non-university level mathematics.
A: This is just an idea, why don't you view it like this
$$ \exp{(A + \frac{1}{2}B + \frac{1}{3}C+ \frac{1}{4}D + \ldots)} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \dots $$
Where $$A = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \dots $$
$$B = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \dots $$
$$C = \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots $$
But this in terms of a function also say $p(n)$ is the function of the $n$th prime
$$ \exp[ \sum_{k=1}^{\infty } ( \frac{1}{k} \cdot    \sum_{n=1}^{\infty} ( \frac{1}{p(n)^k}  ) )] = \sum_{n = 1}^{\infty}  ( \frac{1}{n} ) $$
