limit of product Let $k=n/m \to \infty$ as $n \to \infty$. I am wondering whether we have 
$$\lim_{n\to \infty}\prod_{i=1}^{k}\left(1-\frac{i}{n}\right)=1?$$
I can prove that the limit above exists and equals $1$ under the additional assumption that $n/m^2 \to0$ as $n \to \infty$ by looking at the logarithm of the product and finding a lower bound by using $\log(x)\geq(x-1)/x$. However, I am not sure if that assumption is really necessary.
Edit:
By doing some simulations of the finite product while varying $n$ I am pretty confident that the condition $n/m^2 \to0$ as $n \to \infty$ must be satisfied, I just dont know how to show it is necessary.
 A: As you stated, the condition $n/m^2\to 0$ is also necessary to the convergence of the infinite product to $1$.
More precisely, we have:

Let $k_n\in\Bbb N$ and $1\leq k_n<n$ for every positive integer $n$.
  Then
  $$\prod_{i=0}^{k_n-1}\left(1-\frac{i}{n}\right)\to 1\iff k_n\in o(\sqrt n)$$

We have
\begin{align}
p_n
&=\prod_{i=0}^{k_n-1}\left(1-\frac{i}{n}\right)\\
&=\prod_{i=0}^{k_n-1}\frac{n-i}n\\
&=\frac{n!}{(n-k_n)!n^{k_n}}
\end{align}
By Stirling formula $\log(n!)=n\log(n)-n+O(\log(n))$, we get
\begin{align}
\log(p_n)
&=\log(n!)-\log((n-k_n)!)-k_n\log(n)\\
&=(n\log(n)-n+O(\log(n))-\log((n-k_n)!)-k_n\log(n)\\
&=(n-k_n)\log(n)-n+O(\log(n))-\log((n-k_n)!)\\
&=(n-k_n)\log(n)-n+O(\log(n))-(n-k_n)\log(n-k_n)+(n-k_n)+O(\log(n-k_n))\\
&=(n-k_n)\log(n)-(n-k_n)\log(n-k_n)-k_n+O(\log(n))\\
&=-n[(1-k_n/n)\log(1-k_n/n)+k_n/n]+O(\log(n))\\
&=-nf(k_n/n)+O(\log(n))
\end{align}
where $f(x)=x+(1-x)\log(1-x)$.
Note that $0\leq f(x)\leq 1$ is stricly increasing for $0\leq x\leq 1$ and $f(x)\sim x^2/2$ as $x\to 0$.
If $k_n/n\to\ell>0$ for $n$ in an unbounded subset $S\subseteq\Bbb N$, then $\log(p_n)\sim -nf(\ell)\to-\infty$ as $S\ni n\to\infty$.
Consequently, if $p_n\to 1$, then $k_n/n\to 0$ hence $nf(k_n/n)\sim k_n^2/(2n)$.
By Stirling formula $\log(n!)=n\log(n)-n+\frac 12\log(2\pi n)+O(\log(n))$, we get
\begin{align}
\log(p_n)
&=\log(n!)-\log((n-k_n)!)-k_n\log(n)\\
&=n\log(n)-n+\frac 12\log(2\pi n)+O(1/n)-\log((n-k_n)!)-k_n\log(n)\\
&=(n-k_n)\log(n)-n+\frac 12\log(2\pi n)-\log((n-k_n)!)+O(1/n)\\
&=(n-k_n)\log(n)-n+\frac 12\log(2\pi n)-(n-k_n)\log(n-k_n)+(n-k_n)-\frac 12\log(2\pi(n-k_n)))+O\left(\frac 1{n-k_n}\right)+O\left(\frac 1n\right)\\
&=(n-k_n)\log(n)-(n-k_n)\log(n-k_n)-k_n+\frac 12\log(2\pi n)-\frac 12\log(2\pi(n-k_n)))+O\left(\frac 1n\frac 1{1-k_n/n}\right)+O\left(\frac 1n\right)\\
&=-nf(k_n/n)-\frac 12\log(1-k_n/n)+O\left(\frac 1n\right)\\
\end{align}
hence, we have $\log(p_n)\to 0$ if and only if $k_n^2/(2n)\sim nf(k_n/n)\to 0$.
Consequently, $p_n\to 1$ if and only if $k_n\in o(\sqrt n)$.

To answer your question, let $m_n=n/k_n$.
Then $\frac n{m_n^2}=\frac{k_n^2}n$, hence
$$\prod_{i=0}^{k_n-1}\left(1-\frac{i}{n}\right)\to 1\iff n/m_n^2\to 0$$
