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 $n+1$?