Exponential Function as an Infinite Product 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.
 A: 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,
$$e^x=\prod_{n=1}^\infty\left(\frac{e_n+1}{e_n}\right)^x.$$
A: 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. 
A: Amazingly, the exponential function can be represented as an infinite product of a product! That result was shown in the 2006 paper "Double Integrals and Infinite Products For Some Classical Constants Via Analytic Continuations of Lerch's Transendent" by Jesus Guillera and Jonathan Sondow.
It is proven in Theorem 5.3 that $$e^x=\prod_{n=1}^\infty \left(\prod_{k=1}^n (kx+1)^{(-1)^{k+1} {{n}\choose{k}}}\right) ^{1/n}$$
I dunno, this was too cool not to show you.
A: This function is related to the one Pixel posted, but it is not the same:
$$e=\prod\limits_{n=1}^\infty\left(1-\frac{1}{\tau^n}\right)^{\frac{\mu(n)-\phi(n)}{n}}$$
where $\tau$ denotes the golden ratio, $\mu(n)$ denotes the Möbius Function and $\phi(n)$ denotes Euler's Totient Function.
("A Golden Product Identity for e" by Robert P. Schneider)
A: From Euler's $$\sin[2x]=\Im[e^{xi+xi}]=\Im[[\cos x+i\sin x][\cos x+i\sin x]]=2\sin[x]\cos[x]$$
$$\frac{\sin[πx]}{πx}
=\prod_{n=1}^\infty\left[1^2-\left[\frac{x}n\right]^2\right]$$
we get $$\cos\left[\frac\pi2x\right]=\frac{\sin[πx]}{2\sin[\frac\pi2x]}
=\frac{πx\prod_{n=1}^\infty\left[1^2-[\frac{x}{n}]^2\right]}{2\frac\pi2x\prod_{n=1}^\infty\left[1^2-[\frac{x}{2n}]^2\right]}
\\=\frac{
[\frac{1-x²}{1}][\frac{4-x²}{4}]
[\frac{9-x²}{9}][\frac{16-x²}{16}]\cdots}
{[\frac{4-x²}{4}][\frac{16-x²}{16}]
[\frac{36-x²}{36}][\frac{64-x²}{64}]\cdots}
=\prod_{n=1,3,5}^\infty\left[1^2-\left[\frac{x}{n}\right]^2
\right]$$
which together with$$\sin\left[\fracπ2x\right]=\cos\left[\fracπ2[x-1]\right]$$
yields
$$i^x=\exp\left[i\fracπ2x\right]
=\cos\left[\frac\pi2x\right]+i\sin\left[\frac\pi2x\right]
\\=\left[\prod_{n=1,3,5}^\infty\left[1^2-\left[\frac{x}{n}\right]^2
\right]\right]
+i\left[\prod_{n=1,3,5}^\infty\left[1^2-\left[\frac{x-1}{n}\right]^2
\right]\right]
$$
consisting of two infinite products. Maybe can we factor this expression in some way?
A: For any  $x\in \mathbb{C}/{\mathbb{N}^{-}}$, we have :
$$e^{x}=(1+x)\prod_{n=1}^{\infty}\left(1+\frac{x}{n}\right)^{-n}\left(1+\frac{x}{n+1}\right)^{n+1}$$
A: 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.
A: Not sure if this satisfies your assumptions, but this is an interesting infinite product for $|z|<1$,
$$e^z=\prod_{k=1}^\infty (1-z^k)^{-\frac{\mu(k)}{k}},$$
where $\mu(k)$ is the Möbius function. See here.
Proof
Let's start with the logarithm of the product,
$$-\sum_{k=1}^\infty\frac{\mu(k)}{k}\log\left(1-z^k\right)=\sum_{k=1}^\infty\frac{\mu(k)}{k}\sum_{\ell=1}^\infty\frac{z^{k\ell}}{\ell}.$$
We can rewrite this
$$\sum_{n=1}^\infty\frac{z^n}{n}\sum_{d\mid n}\mu(d)=z+\sum_{n=2}^\infty\frac{z^n}{n}\sum_{d\mid n}\mu(d)=z$$
since $\sum_{d\mid n}\mu(d)=0$ for $n\geq 2$.
A: OP here. I just realized that the following should hold in general:
$$\lim_{n\to\infty}a_n=a_1+\sum_{n=1}^\infty(a_{n+1}-{a_n}),$$
and for finite $a_n$, similarly
$$\lim_{n\to\infty}a_n=a_1\cdot\prod_{n=1}^\infty\frac{a_{n+1}}{a_n}.$$
Hence, with $a_1=\prod_{n=1}^\infty (a_1)^{2^{-n}}$ and for $x$ that aren't negative integers,
$$\mathrm {exp}(x)=\prod_{n=1}^\infty\ (1+x)^{2^{-n}}\left(1+\frac{x}{n+1}\right)^{n+1}\left(1+\frac{x}{n}\right)^{-n}.$$
