TITLE: A product approximation to the Taylor series of the exponential
QUESTION [24 upvotes]: I recently came across the following in something I'm working on, and I'd never seen it before.  Consider
\begin{align*}
f_1(x) &= (1+x)^{1/1}  \\\
f_2(x) &= (1+x)^{2/1} (1+2x)^{-1/2} \\\
f_3(x) &= (1+x)^{3/1} (1+2x)^{-3/2} (1+3x)^{1/3} \\\
f_4(x) &= (1+x)^{4/1} (1+2x)^{-6/2} (1+3x)^{4/3} (1+4x)^{-1/4} \\\
& \cdots \\\
f_n(x) &= \prod_{k=1}^n (1+kx)^{(-1)^{k-1}\binom{n}{k}/k}.
\end{align*}
Think of these as formal power series with coefficients in $\mathbb{Q}$.  It turns out that these series approximate the Taylor expansion for the exponential, in the sense that 
$$ f_n(x) = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} +O(x^{n+1}).$$
(This isn't too hard to prove, using a logarithmic derivative:
\begin{align*}
1-f_n'(x)/f_n(x) &= \sum_{k=0}^n (-1)^k\binom{n}{k}(1+kx)^{-1}
\\\
&= \sum_j \biggl(\sum_k (-1)^k\binom{n}{k}k^j\biggr) (-x)^j.
\end{align*}
The coefficient of $x^j$ calculates the number of surjections $\{1,\dots,j\}\to \{1,\dots,n\}$ (up to a sign), and so is $0$ if $j<n$.)
My questions include (but are not limited to): Do these things have a name?   Is there some kind of combinatorial interpretation of the $f_n(x)$ (e.g., as a generating function of some kind)?  Is there a more direct proof that $f_n(x)$ agrees with $e^x$ to order $n+1$?

REPLY [4 votes]: See A019538, A049019, and A133314 for extensive formula for the coefficients and relation to face vectors of permutahedra/permutohedra. Your proof is a nice one in that A133314 clearly shows why the lower order coefficients as given by the finite differences must vanish if $1/f_n(x)$ truly is an approximation of $exp(-x)$.