Recently I've been toying around with the Totient function and the Prime Number Theorem and came up with the odd result that the following limit
$$\lim_{n\to\infty}\frac{\pi(n)m_n}{\phi(m_n)n}$$
where
$$m_n\equiv\prod_{i=1}^{n}p_i$$
where $p_i$ is the $i$th prime converges to approximately $e$ (somewhat verified computationally).
This is, by the prime number theorem, equivalent to the statement that
$$\lim_{n\to\infty}\prod_{i=1}^{n}\left(\frac{p_i}{p_i-1}\right)/\log n\approx e$$.
Or, taking the logarithm,
$$\lim_{n\to\infty}\left[\sum_{i=1}^{n}\log\left(\frac{p_i}{p_i-1}\right)-\log\log n\right]\approx 1$$
Furthermore, by a relation involving the Euler-Mascheroni constant, this implies
$$\lim_{n\to\infty}\sum_{n<p\leq p_n}\log\left(\frac{p}{p-1}\right)\approx 1-\gamma$$
My question is whether or not anyone knows how to prove any of these statements, either with equality (perhaps computation is leading me astray) or with a different value as the limit.
Some of your formula's are not quite correct. It is a result of Mertens that $$\prod_{p\leq x} \left(1-\frac{1}{p}\right)^{-1} \sim e^\gamma \log x,$$ (see Merten's Third Theorem on this page) and we are taking $x=p_n \sim n\log n$ so it follows that
$$\lim_{n\rightarrow \infty} \frac{n}{\phi(n)} \frac{\pi(n)}{n}=e^\gamma.$$ Consequently
$$\lim_{n\rightarrow \infty} \left[ \sum_{i=1}^n \log\left(\frac{p_i}{p_i-1}\right)-\log \log n\right]=\gamma.$$
Now, using the fact that $$\sum_{p\leq x}\log\left(\frac{p}{p-1}\right)\sim\log\log x+\gamma$$ and $p_n\sim n\log n$ we find that $$\sum_{n<p\leq p_{n}}\log\left(\frac{p}{p-1}\right)\sim 0.$$