TITLE: Exponential Function as an Infinite Product
QUESTION [21 upvotes]: Is there any representation of the exponential function as an infinite product (where there is no maximal factor in the series of terms which essentially contributes)? I.e.
$$\mathrm e^x=\prod_{n=0}^\infty a_n,$$
and by the sentence in brackets I mean that the $a_n$'s are not just mostly equal to $1$ or pairwise canceling away. The product is infinite but its factors don't contain a subseqeunce of $1$, if that makes sense.
There is of course the limit definition as powers of $(1+x/n)$., but these are no definite $a_n$'s, which one could e.g. divide out.

REPLY [2 votes]: We have$$\mathrm e^{-x}=\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$$$(1-x)(1+\tfrac12x^2)(1+\tfrac13x^3)(1+\tfrac38x^4)(1+\tfrac15x^5)(1+\tfrac{13}{72}x^6)(1+\tfrac17x^7)(1+\tfrac{27}{128}x^8)(1+\tfrac{8}{81}x^9)\cdots$$for $|x|<1$. Some coefficients for these product terms are listed in OEIS as A170910 and A170911. See the paper on power product expansions by Gingold et al. (1988). The result is due to O. Kolberg (1960), who showed that the coefficient of $x^n$ in the $n$th factor is $1/n$ if and only if $n$ is prime.