Approximating the exponential function I have found experimentally something that seems graphically like an approximation of the exponential function. However, it is totally experimental and I have no idea whether it really converges towards the $\exp$.
Let : $$f\left(x,h,c\right)=\left(1+\frac{x}{c^h}\right)^{c^h}\text{(A quite understandable approximation for the exponential function)}$$
$$q\left(x,h,c\right)=\sum \limits_{p=0}^h\frac{c^{\frac{p^2+p}{2}}}{\left(\prod\limits _{i=1}^p\left(c^i-1\right)\right)\prod\limits _{i=1}^{h-p}\left(1-c^i\right)}f\left(x,p,c\right)$$
My approximation is : $$\exp(x)=\underset{h,c\rightarrow+\infty}{\lim}q(x,h,c)$$
Desmos shows that it s indeed really close near $1$, and that for low $h,c$ it starts diverging afterwards (but this might be because of computational errors on huge numbers (?) ).
Is that a known approximation for the exponential ?
If not, is it an approximation of the exponential at all ?
Additional question that sparked from the comments : How can the comportment for low $h,c$ be analyzed ?

If you wonder where that formula comes from, I can't give a full explanation (it's really experimental work) but you might be interested by one of my previous questions An expression for $U_{h,0}$ given $U_{n,k}=\frac{c^n}{c^n-1}(U_{n-1,k+1})-\frac{1}{c^n-1}(U_{n-1,k})$.
 A: As shown in equation $(6)$ from this answer,
$$
\prod_{k=1}^h\frac{c^kx-1}{c^k-1}=\sum_{p=0}^h\frac{\displaystyle c^{p(p+1)/2}}{\displaystyle\prod_{k=1}^p(c^k-1)\prod_{k=1}^{h-p}(1-c^k)}x^p\tag{1}
$$
Plugging $x=1$ into $(1)$, we get that
$$
\sum_{p=0}^h\frac{\displaystyle c^{p(p+1)/2}}{\displaystyle\prod_{k=1}^p(c^k-1)\prod_{k=1}^{h-p}(1-c^k)}=1\tag{2}
$$
For $x\ge0$, $0\le e^x-\left(1+\frac xn\right)^n\le\dfrac{x^2e^x}{2n}$. Therefore,
$$
\begin{align}
|e^x-f(x,p,c)|
&=\left|\,e^x-\left(1+\frac{x}{c^p}\right)^{c^p}\,\right|\\
&\le\frac{x^2e^x}{2}c^{-p}\tag{3}
\end{align}
$$
Since the coefficients of $x^p$ in $(1)$ alternate sign, we get
$$
\begin{align}
&\left|\,e^x-\sum_{p=0}^h\frac{\displaystyle c^{p(p+1)/2}}{\displaystyle\prod_{k=1}^p(c^k-1)\prod_{k=1}^{h-p}(1-c^k)}f(x,p,c)\,\right|\\[6pt]
&=\left|\,\sum_{p=0}^h\frac{\displaystyle c^{p(p+1)/2}}{\displaystyle\prod_{k=1}^p(c^k-1)\prod_{k=1}^{h-p}(1-c^k)}(e^x-f(x,p,c))\,\right|\tag{4}\\[6pt]
&\le\frac{x^2e^x}{2}\left|\,\sum_{p=0}^h\frac{\displaystyle c^{p(p+1)/2}}{\displaystyle\prod_{k=1}^p(c^k-1)\prod_{k=1}^{h-p}(1-c^k)}\left(-\frac1c\right)^p\,\right|\tag{5}\\[18pt]
&=\frac{x^2e^x}{2}\prod_{k=1}^h\frac{c^{k-1}+1}{c^k-1}\tag{6}
\end{align}
$$
Explanation:
$(4)$: apply $(2)$
$(5)$: use $(3)$
$(6)$: apply $(1)$
For both a fixed $c\gt1$ as $h\to\infty$ and a fixed $h\ge1$ as $c\to\infty$, $(6)$ vanishes.
