I'm trying to prove that the the first derivative of the exponential generating function of the Harmonic numbers, $H_n$, is the exponential generating function of $H_{n+1}$, but I my solution seems flawed:
for $H_n = \sum_0^\infty \frac{1}{x}$
the EGF $H(z)= \sum_{n=0}^\infty H_n \frac{z^n}{n!}$.
\begin{align*} Then\, H'(z) = \frac{dH(z)}{dz} &= \sum_{n=0}^\infty H_nn\frac{z^{n-1}}{n!} \tag{ peform differentiation}\\ &= \sum_{n=0}^\infty H_n\frac{z^{n-1}}{(n-1)!} \tag{cancel $n$ against the $n$ in the $n!$}\\ &= \sum_{i=-1}^\infty H_i \frac{z^i}{i!} \tag { let $i=n-1$, change index}\\ &= \sum_{i=0}^\infty H_{i+1} \frac{z^i}{i!} \tag{can i just change the index from $i=-1$ to $i=0$? why?}\\ &= \text{the EGF of $H_{n+1}$}\\ \end{align*}
Is this correct?