How to prove this identity on exponential generating function of harmonic numbers I came across the following problem, let $N![z^N]A(z)$ denote the coefficient of an exponential generating function (EGF) $A(z)$. The EGF is similar to an ordinary generating function (OGF) $A'(z)$ except that instead of the series $A'(z)=\sum_0^Na_Nz^N$ for an OGF, we have $A(z)=a_Nz^N/N!$ for an EGF $A(z)$.
For instance, for EGF $A(z)=e^z$, we have $N![z^N]A(z)=N![z^N]e^z=1$, i.e. the EGF coefficients of $e^z$ are $1$ for all $N \in \mathbb{N}$, i.e. the coefficients of the set $ \{z^0/0!,z^1/1!,z^2/2!,...,z^N/N! \}$ are all $1$ for any $N$ given $e^z$. Similarly, for EGF $A(z)=1/(1-z)$, we have $N!z[^N](1/(1-z))=N!$.
Now, given the following EGF $A(z)$:
$$
A(z)=e^z\int^z_0\frac{1-e^{-t}}{t}dt
$$
We are supposed to get $N![z^N]A(z)=H_N$, where $H_N$ is the $N$th harmonic number, i.e.
$$
N![z^N]e^z\int^z_0\frac{1-e^{-t}}{t}dt = H_N
$$
I could not think of a way to prove the above statement. The problem gave a hint that proving this statement involves forming a differential equation for the EGF $H(z)=\sum_{N \geq 0}H_Nz^N/N!$...
Any help?
 A: I think "forming a differential equation" is an overkill. We can simply do
$$A(z)=\int_0^z\frac{e^z-e^{z-t}}{t}\,dt\underset{t=z(1-x)}{\phantom{\big[}=\phantom{\big]}}\int_0^1\frac{e^z-e^{zx}}{1-x}\,dx=\sum_{n=1}^{\infty}\frac{z^n}{n!}\int_0^1\frac{1-x^n}{1-x}\,dx=\sum_{n=1}^{\infty}H_n\frac{z^n}{n!}$$ (the last equality, if unknown to you, follows from $(1-x^n)/(1-x)=1+\ldots+x^{n-1}$).
A side note: if we directly multiply the series for $e^z$ and the integral, we get $$A(z)=\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\left(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}z^n}{n\cdot n!}\right)=\sum_{n=1}^{\infty}\frac{z^n}{n!}\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{k-1}}{k},$$ i.e. another proof of the "frequent" $G_1=H_1$ case of this question of mine.
A: Here is a proof involving forming a differential equation.

Let $H_n$ be the $n^\text{th}$ harmonic number. In particular, $H_0=0$ and $H_n=\frac1n+H_{n-1}$ for $n\geq 1$.
Define the series $B(t)=\sum_{n\geq 0} H_{n+1}\frac{t^n}{n!}$.
Let $A(z) = \int_0^z B(t)dt$.
One observation is that:
$$A(t)=\sum_{n\geq 0} H_n\frac{t^n}{n!}\label{1} \tag{1}.$$
With that in mind:
\begin{align*}
  B(t)
  &= \sum_{n\geq 0} H_{n+1}\frac{t^n}{n!} \\
  &= \sum_{n\geq 0}\left(\frac1{n+1} + H_n\right)\frac{t^n}{n!} \\
  &= t^{-1}\sum_{n\geq 1}\frac{t^n}{n!} + \sum_{n\geq 0}H_n\frac{t^n}{n!} \\
  &= t^{-1}(e^t-1) + \int_0^t B(s)ds. \\
\end{align*}
In other words, $A'(t)-A(t) = t^{-1}(e^t-1)$. This is a $1^\text{st}$ order linear ODE.
By multiplying both sides by the integrating factor $e^{-t}$, we get:
\begin{align*}
[e^{-t}A(t)]' = t^{-1}(1-e^{-t}).
\end{align*}
Integrate both sides from $t=0$ to $t=z$ and rearrange to find:
$$A(z)=e^z\int_0^z \frac{1-e^{-t}}{t}dt.$$
Using (\ref{1}), we can see that $n![z^n]A(z)=H_n$.
