TITLE: Prove that there is $n\in\mathbb{N}$ such that $n!$ starts with 1996
QUESTION [11 upvotes]: Basically the statement says that there is some number $n$ such that:
$$1996\cdot 10^k<n!<1997 \cdot 10^k$$
$$k + \log 1996<\sum_{i=1}^n \log i<k+\log 1997$$
A nice idea how to solve a similar problem can be found here, but I could not utilize it in this problem. 
Origin of this problem is a mystery - I have found it on a page where people listed their favorite math problems but this one had no solution.

REPLY [4 votes]: Examples
Looking at a few examples gives a clue as to why there will be an infinite number of such numbers.
$$
\begin{align}
10003183!&=1.9957478783343368038\times10^{65679340}\\
10003184!&=\color{#C00}{1.996}3833244587984566\times10^{65679347}\\
10003185!&=1.9970191725476385839\times10^{65679354}\\
100006487!&=1.9959743653614076660\times10^{756622452}\\
100006488!&=\color{#C00}{1.996}1038641782323141\times10^{756622460}\\
100006489!&=\color{#C00}{1.996}2333913579788396\times10^{756622468}\\
100006490!&=\color{#C00}{1.996}3629469050779724\times10^{756622476}\\
100006491!&=\color{#C00}{1.996}4925308239615811\times10^{756622484}\\
100006492!&=\color{#C00}{1.996}6221431190626726\times10^{756622492}\\
100006493!&=\color{#C00}{1.996}7517837948153934\times10^{756622500}\\
100006494!&=\color{#C00}{1.996}8814528556550287\times10^{756622508}\\
100006495!&=1.9970111503060180035\times10^{756622516}
\end{align}
$$
In the case of $10003183!$, whose mantissa is a bit small, when we multiply the factorial by $10003184$, we multiply the mantissa by $1.0003184$. Since the mantissa is approximately $2$, it is increased by approximately $0.0006368$, and that brings $1.9957479$ to $1.9963833$.
In the case of $100006487!$, whose mantissa is a bit small, when we multiply the factorial by $100006488$, we multiply the mantissa by $1.00006488$. Since the mantissa is approximately $2$, it is increased by approximately $0.00012976$, and that brings $1.9959743$ to $1.9961039$.

Theoretical Considerations
$\newcommand{\bigO}[1]{\operatorname{O}\left(#1\right)}$
For $n\ge8$ and $m^2\le2\log(10)\,10^n$, we have
$$
\left\{\log_{10}\left(\frac{\left(10^n+m\right)!}{10^n!}\right)\right\}=\frac{m^2}{2\log(10)\,10^n}+\bigO{\frac1{10^{n/2}}}\tag1
$$
where $\{x\}=x-\lfloor x\rfloor$ is the positive fractional part of $x$. So for
$$
m\approx\left(2\log(10)\,10^n\left\{\log_{10}(1996)-\log_{10}\left(10^n!\right)\right\}\right)^{1/2}\tag2
$$
we should have
$$
\left\{\log_{10}\left(\left(10^n+m\right)!\right)\right\}\approx\left\{\log_{10}(1996)\right\}\tag3
$$
If the approximation in $(3)$ is a bit off, we can adjust things by noting that
$$
\begin{align}
\left\{\log_{10}\left(10^n+m\right)\right\}
&\approx\frac{m}{\log(10)\,10^n}\\
&\le\left(\frac2{\log(10)\,10^n}\right)^{1/2}\tag4
\end{align}
$$
which is small enough to adjust within an interval of width $\log_{10}\left(\frac{1997}{1996}\right)\approx\frac1{1996\,\log(10)}$ when $n\ge8$.