Maximizing $(\frac{N}{k})^k$ for integer k - proof For a given integer $N$, we want to maximize $(\frac{N}{k})^k$. The solution in real numbers is obtained when $k = \frac{N}{e}$. The solution in integers is then either $\lfloor\frac{N}{e}\rfloor$ or $\lceil\frac{N}{e}\rceil$. However, it seems that the correct solution is always rounding to the nearest integer.
Is this assumption correct and if so, what's the proof?
 A: What about writing the taylor development to the 3rd derivative ? That's a good place to start in the maximum.
Let's imagine that N/e is closer to the upper integer. You have n < n+1/2 < N/e < n+1.
f' = 0
f'' = something < 0 but it acts symetrically, and so f(n) will be pushed farther from the maximum than f(n+1) which is closer to it N/E
f''' > 0 so this will push f(n) negatively and f(n+1) positively

So in this case it seems true that if N/e is closer to n+1 than to n, then f(n+1) is bigger than f(n)
Second case:
f'' will push f(n) negatively less than f(n+1) but f''' will push f(n) negatively and f(n+1) positively so you are trying to find a place where this effect is bigger than the previous one to find a counter example if any. I haven't done the calculation here but you can come back with questions if any
Note : 1/e is irrational so you can find N such that N/e is arbitrarily close to m+0.5
Note 2: I say "seem" in the first case at the end because a Taylor development is cool but we have to analyze the rest too and minor it.
This is not a full proof but it gives you a good step forward
