Let

\begin{align*}
c_n &= n!\left(e-\sum_{k=0}^n \frac{1}{k!}\right) \\
\\
u_n &= \bigg\lfloor{\frac{1}{c_n} \bigg\rfloor} \\
\\
v_n &= \bigg\lfloor{\frac{1}{1/c_n-\lfloor{u_n} \rfloor}}  \bigg\rfloor
\end{align*}

Are $u_n = n$ and $v_n = n+1$ for all $n \geq 0$? 

The question extends as follows (January 16).  Let

\begin{align*}
r_{1,n} &= 1/c_n\\ 
r_{2,n} &= 1/r_{1,n}\\
r_{3,n} &= 1/r_{2,n}\\
&\vdots\\
s_{1,n} &= \lfloor{1/\{r_{1,n}\}}\rfloor, \text{ where }\{ \} \text{ denotes fractional part}\\
s_{2,n} &= \lfloor{r_{2,n}} \rfloor \\
s_{3,n} &= \lfloor{r_{3,n}} \rfloor \\
&\vdots\\
\end{align*}

Are $(s_{2,n})=(1,1,2,2,3,3, \ldots)$ and $(s_{3,n})=(2,1,3,1,4,1,6,1,7,1, \ldots)$, where the only numbers missing are $5+4h$ for $h \geq0$?
In general, does the regularity of the continued fraction for $e$ imply some sort of regularity for $(s_{k,n})$ for $k \geq 5$?

Informally speaking, things look very irregular for higher $k$, and I wonder if the method in Answer 1 extends to these deeper cases.