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.

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?
