I was playing around with numbers like I always do, and I came up with a conjecture: $$\ln(e^{\pi + n} - e^{\pi + n - 1} - \ldots - e^{\pi + 1} - e^{\pi}) - \frac{\pi}{\sqrt{2}} \approx n$$ such that $n\in\mathbb{N}$. I then realised that the larger the value of $n$, the more accurate the LHS was getting closer to the RHS...
...and then it stopped. Strangely enough, it appeared that $$\ln\left(e^{\pi + n} - \sum_{i = 0}^{n-1}e^{\pi + i}\right) - \frac{\pi}{\sqrt{2}} \sim n + k$$ such that, as far as I have tested, $k \approx 0.0479330616936374724102$ for all $n\leqslant 61$. Subtracting this constant $k$ from both sides, I see that $$\frac{\pi}{\sqrt{2}} + k \approx 2.269374530772820595918141.$$ Now by letting $g = \dfrac{\pi}{\sqrt{2}} + k$, is there some algebraic formula (like an infinite series) to find $k$?
If there is, and we find it, we can find the value of $g$, and we would be able to write down that, $$\ln\left(e^{\pi + n} - \sum_{i=0}^{n-1}e^{\pi + i}\right) - g\sim n$$ with whatever $g$ is equal to (algebraically speaking). For now, $g$ is just a special constant.
Thanks in advance.
Hint Factoring and applying the product-to-sum rule for logarithms gives $$\log(e^{\pi + n} - e^{\pi + n - 1} - \cdots - e^{\pi}) = \pi + n + \log(1 - e^{- 1} - \cdots - e^{-n}) .$$ Now, rewriting the argument of the logarithm on the right-hand-side using the formula for the sum of a geometric sequences gives $$1 - \frac{1 - e^{-n}}{e - 1} .$$
What happens as $n \to \infty$?
Note, by the way, the that there's nothing special about the use of $\pi$ here---we could just as well replace it with any other number.