What are some good bounds for the logarithm of the product of the first $n$ primes greater than $m$?
I also want to address a confusion I have:
The logarithm of the product of the first $n$ primes is roughly $n\log n$ so the product in question should be roughly $$(n+\pi(m))\log(n+\pi(m))-\pi(m)\log\pi(m),$$ but this is smaller than $n\log m$, which is absurd.
On
If k is small compared to n, then the k-th prime after m is about m + k log m or m * (1 + k/m log m). The logarithm is about log m + (k/m log m). The sum for the first k primes after m would be about k log m + (k^2 / 2m * log m), and the value about m^k * m^(k^2 / 2m).
A comment asked about k = 100, $m = 10^{15}$, here the product is $m^{1500} * (10^{15})^{5000 / 10^{15}}$ or $m^{1500} * (1 + 1.727 / 10^{10})$ apart from mistakes I may have made.
This is an estimate, not bounds. For $k=100$ and $n > 100000$ it should be reasonably accurate.
We have the well known formula for the sum of logarithm of primes up to $x$ $$ \sum_{p \leq x}\log p = \Theta(x), $$ from which it follows that $$ \sum_{m \leq p \leq x} \log p = O(x-m). $$
The $n$'th prime is $\Theta(n \log n)$ so $$ \sum_{i \leq n}\log p_i \approx \sum_{p \leq n\log n} \log p = \Theta(n \log n). $$ The first prime after $m$ is $\pi(m)$'th prime, so the last prime is the $\pi(m) + n$'th prime, therefore would be of size $\Theta\left((\pi(m) + n)\log(\pi(m) + n)\right)$.
the result that you are looking for is $$ \sum_{m \leq p \leq (\pi(m) + n)\log(\pi(m) + n)} \log p = O((\pi(m) + n)\log(\pi(m) + n)-m). $$