Function satisfying $f(z) = 1 + z f\left(\frac{z}{1+z}\right)$ 
Iterate the following equation to obtain an explicit formula for $f( z)$:
$$
\begin{align*}
f( z) = 1 + z f\left( \frac{z}{1 + z}\right)
.\end{align*}
$$

Iterating this equation one obtains
$$
\begin{align*}
f( z) = \sum_{n \geq 0}^{} \frac{z^{n}}{\prod_{k = 0}^{n - 1} ( 1 + kz)}
.\end{align*}
$$
I wonder know whether this is already "explicit" enough or whether it can be simplified
even further.

Substituting $z\mapsto\frac1z$ in the functional equation and iterating yields
$$f(1/z)=1+(1/z)f(1/(z+1))=\sum_{n=0}^\infty\frac1{z^{\overline n}}$$
which DLMF 8.11.4 simplifies to
$$f(z)=e\gamma(1/z,1)+1$$
where $\gamma$ is the lower incomplete gamma function. $f$ has simple poles at the negative reciprocals of the positive integers; this can be seen by attempting to expand the functional equation at those points.
Taking the one-sided derivatives at $0$ from above reveals that $f$ is the ordinary generating function for a shifted, sign-alternated version of the Rao Uppuluri–Carpenter numbers A000587:
$$f(z)=1+z+z^2-z^4+z^5+2z^6-9z^7+9z^8+50z^9-267z^{10}+\cdots$$
