Exponential Function as an Infinite Product

Question

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.

Answer 1

If $x\geqslant0$ (or $x\ne-2^n$ for every $n\geqslant0$), one can use $$a_0=1+x,\qquad a_{n+1}=\left(1+\frac{x^2}{2^{n+2}(x+2^n)}\right)^{2^n} $$ If $x\leqslant0$ (or $x\ne2^n$ for every $n\geqslant0$), one can use $$a_0=\frac1{1-x},\qquad a_{n+1}=\left(1-\frac{x^2}{(2^{n+1}-x)^2}\right)^{2^n} $$ Where does this come from? From the identity, valid for every $n\geqslant0$, $$ \prod_{k=0}^na_k=\left(1\pm\frac{x}{2^n}\right)^{\pm2^n}. $$ The first identity (when $\pm=+$) yields a nondecreasing sequence of partial products. The second identity (when $\pm=-$) yields a nonincreasing sequence of partial products.

Answer 2

There exists an infinite product for $e$ as follows:

If we define a sequence $\lbrace e_n\rbrace$ by $e_1=1$ and $e_{n+1}=(n+1)(e_n+1)$ for $n=1,2,3,...;$ e.g. $$e_1=1,e_2=4,e_3=15,e_4=64,e_5=325,e_6=1956,...$$ then $$e=\prod_{n=1}^\infty\frac{e_n+1}{e_n}=\frac{2}{1}.\frac{5}{4}.\frac{16}{15}.\frac{65}{64}.\frac{326}{325}.\frac{1957}{1956}. ...$$ For proof, first by induction we can show that if $s_n=\sum_{k=0}^n\frac1 {k!}$, then $e_n=n!s_{n-1}$,for $n\in\mathbb N$. And this immediately follows that $s_n/s_{n-1}=(e_n+1)/e_n$ and $s_n=\prod_{k=1}^n\frac{e_k+1}{e_k}$. Hence, $$e^x=\prod_{n=1}^\infty\left(\frac{e_n+1}{e_n}\right)^x.$$