Exponential Function as an Infinite Product

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 $n\geq 2$.

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.

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, $$e^x=\prod_{n=1}^\infty\left(\frac{e_n+1}{e_n}\right)^x.$$