Proving an heuristic bound involving the gaussian Consider the pretty sum $S_n=\sum_{k=1}^{n} \prod_{i=0}^{k-1} (1-\frac{i}{n})$. Using the first order approximation of the convex function $e^{-x}$ around 0, we find that $1-x \sim e^{-x}$, so the product is expected to behave like:
$$\sum_{k=1}^{n} e^{-\sum_{i=0}^{k-1} i/n} \sim \sum_{k=1}^{n}e^{-\frac{1}{2} (k/\sqrt{n})^2}\sim \sqrt{n} \int_{0}^{\sqrt{n}}e^{-x^2/2}dx\sim \sqrt{n \pi/2}$$
But I find it hard to justify the first approximation. 


*

*My first goal is to show $S_n \ge C \sqrt{n}$ for some $C$. I know it will follow by showing that $\prod_{i=1}^{\sqrt{n}} (1-\frac{i}{n})$ is bounded from below by an absolute positive constant. I can show it using Stirling's Approximation, but I want to avoid it.

*My second goal is to actually prove $S_n \sim \sqrt{n \pi/2}$ (I am not even sure it is true).
Motivation: This sum appeared while trying to prove Stirling's approximation in an elementary fashion.
Edit: I solved the first question, by taking logarithm and using the 2nd order Taylor approximation of $\ln(1-x)$ around 0.
 A: Using first $\log(1-i/n)\leqslant-i/n$ for every $i\leqslant k-1$ and $k\leqslant n-1$, and then 
$$
\sum_{i=0}^{k-1}\frac{i}n=\frac{k(k-1)}{2n}\geqslant\frac{x^2}2,
$$
for every $k\geqslant2$ and every $x$ in $((k-2)/\sqrt{n},(k-1)/\sqrt{n})$, one gets
$$
S_n\leqslant\sum_{k=1}^{n-1}\exp\left(-\frac{k(k-1)}{2n}\right)\leqslant1+\sqrt{n}\int_0^{(n-2)/\sqrt{n}}\mathrm e^{-x^2/2}\mathrm dx,
$$
hence, for every $n\geqslant1$,
$$
S_n\leqslant1+\sqrt{n}\int_0^{\infty}\mathrm e^{-x^2/2}\mathrm dx=1+\sqrt{n\pi/2}.
$$
Likewise, $\log(1-x)\leqslant-x-x^2$ for every $x$ in $(0,1/2)$ hence $\log(1-i/n)\geqslant-i/n-(i/n)^2$ for every $i\leqslant k-1$ and $k\leqslant n^a$, where $a\lt1$, for every $n$ large enough (namely, such that $n\gt2n^a$). Hence, 
$$
\sum_{i=0}^{k-1}\frac{i}n+\left(\frac{i}n\right)^2\leqslant\frac{x^2}2+\frac{x^3}{3\sqrt{n}}\leqslant\frac{x^2}2+\frac{n^{3a-2}}{3},
$$
for every $k\leqslant n^a$ and every $x$ in $((k-1)/\sqrt{n},k/\sqrt{n})$, hence
$$
S_n\geqslant\sqrt{n}\mathrm e^{-n^{3a-2}/3}\int_0^{n^{a-1/2}}\mathrm e^{-x^2/2}\mathrm dx,
$$
For every $a\gt2/3$, the prefactor is equivalent to $\sqrt{n}$. For every $a\gt1/2$, the integral converges to $\sqrt{\pi/2}$. Hence the result follows, choosing $a=5/12$ for example.
A: As a follow up to my comment let's show the first terms of the expansion of your sum :
 $$\displaystyle S(n)=\sqrt{n\frac{\pi}2}-\frac 13+\frac 1{12}\sqrt{\frac {\pi}{2\,n}}-\frac 4{135\,n}+\frac 1{2\cdot 12^2}\sqrt{\frac {\pi}{2\,n^3}}\\+\frac 8{2835\,n^2}-\frac{139}{30\cdot 12^3}\sqrt{\frac {\pi}{2\,n^5}}+\frac {16}{8505\,n^3}-\frac{571}{120\cdot 12^4}\sqrt{\frac {\pi}{2\,n^7}}+O\left(n^{-4}\right)$$
But Ramanujan got (OEIS) :
$$\int_0^\infty e^{-x}\left(1+\frac xn\right)^n dx=\frac{e^n\;n!}{2\;n^n}+\frac 23-\frac 4{135\,n}+\frac 8{2835\,n^2}+\frac {16}{8505\,n^3}-\frac {8992}{12629925\,n^4}-\cdots$$
or the alternative form :
$$\sum_{k=0}^{n-1} \frac {n^k}{k!} + \left(\frac 13+\frac 4{135\,n}-\frac 8{2835\,n^2}-\cdots\right)\frac {n^n}{n!} = \frac {e^n}2$$
And the $\sqrt{\frac{\pi}2}$ part of my expansion is merely a $\Gamma$ function expansion :
$$\Gamma(x)\sim \sqrt{2\pi}\;x^{x-1/2}  e^{-x}\left(1+\frac 1{12x}+\frac 1{288x^2}-\frac{139}{51840x^3}-\cdots\right)$$
All this is clearly related to the Stirling formula so that Berndt's 'Ramanujan's notebooks II' or the Question $294$ may interest you (both containing further references).
I'll let you put all this together and wish at least some entertainment,
