Asymptotic equivalence of exponents An earlier question links to a paper of Erdos in which he says that it is "well-known" that the Prime Number Theorem is equivalent to 
$(\prod_{p\leq n}p)^{1/n} \to e$ as $n\to \infty.$ **
Here is my confusion. 
If $~\prod_{p\leq n}p \sim e^n$ or $e^{\log \prod p}= e^{\sum \log p} \sim e^n,$
(the last relation appears in the linked question, but I take responsibility for  it) doesn't this imply that 
(*) $\lim_{n \to \infty} (\sum_{p\leq n} \log p - n) = 0?$
Of course it's true that $\lim_{n \to \infty} \frac{\sum \log p}{n} =1 $ and I do not think that (*) is true. But I think we do have in general that 
$$ e^{f(x)}\sim e^{g(x)} \implies \lim (f(x) - g(x)) = 0,$$ since $\lim \frac{e^f}{e^g}= e^{f-g} = 1 \implies \lim (f-g) = 0.$ 
Can someone tell me where I have goofed? Thanks!
**If someone could point me to a proof of this I would appreciate it --I don't see it in Apostol or Hardy & Wright).

$(\Pi_{ p \leq n} p)^{1/n} \rightarrow e$, as $n \rightarrow \infty$, doesn't imply $\Pi_{p \leq n} p \sim e^n$.
Example: $(2^n n)^{1/n}  \rightarrow 2$, as $n\rightarrow \infty$, but $2^n n$ is not equivalent to $2^n$.
Indeed, $n^{1/n}=e^{\frac{\log n}{n}} \rightarrow 1$ because $\frac{\log n}{n} \rightarrow 0$ as $n \rightarrow \infty$
