Is this a known formula? $ \prod_{k=0}^n \left(1 - \frac{a_k}{N}\right)$ I try to quantify a partition, are there any known indicators/caracteristic numbers? Something which came to my mind was
$$ \prod_{k=0}^n \left(1 - \frac{a_k}{N}\right), $$ 
with the following condition
$$ \sum_{k=0}^n a_k = N .$$
Is this a known formula? I'd like to have an indicator which tells me if the partition is well spread or concentrated on some few numbers.
I hope my question is clear as I know not a lot about partitions. Thank you for your help. 
Edit: If $a_0=N$ and all others $a_k$ are 0 this formula gives 0. If all $a_k$ are 1 and $n=N$ tends to infinity this formula goes to $\frac{1}{e}$. So I am wondering too if $\frac{1}{e}$ is the upper bound for a finite $N$ for all partitions.
Edit2: Thanks a lot for different proofs that $\frac{1}{e}$ is the upper bound.
I still like to know if someone knows something more about this formula. If someone has an interessing fact, that would be nice.
 A: Let $f(x) = -\log(1-x)$ for $x \in [0,1)$. 
Since $f''(x) = \frac{1}{(1-x)^2} > 0$. $f(x)$ is convex over $[0,1)$.
By Jensen's inequality,
for any $a_0,\ldots, a_n \ge 0$ such that $\sum_{i=0}^n a_i = N$, we have
$$-\sum_{i=0}^n\log\left(1 - \frac{a_i}{N}\right) \ge -(n+1)\log\left(1-\frac{1}{n+1}\right)$$
In the Taylor expansion of $f(x) = \sum\limits_{k=1}^\infty \frac{x^k}{k}$, all coefficients are positive. This leads to
$$-(n+1)\log\left(1-\frac{1}{n+1}\right) = (n+1)\sum_{k=1}^\infty \frac{(n+1)^{-k}}{k} >
(n+1)\sum_{k=1}^1 \frac{(n+1)^{-k}}{k} = 1$$
As a result,
$$\prod_{i=0}^n\left(1 - \frac{a_i}{N}\right) \le \left(1-\frac{1}{n+1}\right)^{n+1} < e^{-1}$$
In short, $e^{-1}$ is indeed the upper limit for a finite $N$ for all partitions.
A: Since $$1-x\leq e^{-x},$$ if $x\geq 0,$ the upper bound you state:
$$ \prod_{k=0}^n \left(1 - \frac{a_k}{N}\right)\leq e^{-a_1/N} \times \cdots\times e^{-a_N/N} =e^{-1}, $$ 
holds for any finite partition.
Your measure can be looked at as the $n^{th}$ power of the geometric mean of the relative sizes of the complements of partition atoms.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{\ds{\bracks{\prod_{k = 0}^{n}\pars{1 - {a_{k} \over N}}}
\bracks{z^{N}}z^{\sum_{j = 0}^{n}a_{j}}}} =
\bracks{z^{N}}\prod_{k = 0}^{n}\bracks{\pars{1 - {a_{k} \over N}}z^{a_{k}}}
\\[1cm]
&\ln\pars{\prod_{k = 0}^{n}\bracks{\pars{1 - {a_{k} \over N}}z^{a_{k}}}} =
\sum_{k = 0}^{n}\bracks{\ln\pars{1 - {a_{k} \over N}} + a_{k}\ln\pars{z}}
\\[5mm]\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,&
\bracks{\ln\pars{z} - {1 \over N}}\sum_{k = 0}^{n}a_{k}
\\[1cm] &\
\prod_{k = 0}^{n}\bracks{\pars{1 - {a_{k} \over N}}z^{a_{k}}}
\,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,
z^{\sum_{j = 0}^{n}a_{j}}\expo{-\pars{\sum_{j = 0}^{n}a_{j}}/N} = z^{N}\expo{-1}
\\[1cm]
&\bbox[10px,#ffd]{\ds{\bracks{\prod_{k = 0}^{n}\pars{1 - {a_{k} \over N}}}
\bracks{z^{N}}z^{\sum_{j = 0}^{n}a_{j}}}}
\,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\, \bbx{\expo{-1}}
\end{align}
