The following exercise is out of Stanley's Enumerative Combinatorics: Vol. II (Ex. 7.63):

For $\lambda \vdash n$ define $d_\lambda = \sum_{w \in D_n} \chi^\lambda(w)$ where $D_n$ is the set of all derangements in $S_n$. Show that

$\sum_{\lambda \vdash n} d_\lambda s_\lambda = \sum^n_{k=0}(-1)^{n-k}(n)_k h_{1^{n-k}k}$

To prove this, Stanley suggests using Murnaghan-Nakayama to arrive at $d_\lambda = n!s_\lambda |_{p_1=0,p_2=p_3= \cdots =0}$. Then by the Cauchy Identity and Proposition 7.7.4 in Stanley Vol. II: $\prod_{i,j}(1-x_iy_j)^{-1} = \sum_\lambda s_\lambda(x)s_\lambda(y) = \exp \sum_{n \geq 1}\frac{1}{n}p_n(x)p_n(y)$

Setting $p_1(y) = 0$ and $p_2(y) = p_3(y) = \cdots = 1$ we have:

$\sum_\lambda \frac{d_\lambda}{n!}s_\lambda(x) = \exp \sum_{n \geq 2}\frac{1}{n}p_n(x)$

Using the identity $\exp \sum_{n \geq 1}\frac{1}{n}p_n(x)p_n(y) = \sum_\lambda z_\lambda^{-1}p_\lambda(x)p_\lambda(y)$ and the fact that $h_n = \sum_{\lambda \vdash n}z_\lambda^{-1}p_\lambda$, it seems to me that we have $\exp \sum_{n \geq 2}\frac{1}{n}p_n(x) = e^{-h_1} \sum_{\lambda}z_\lambda^{-1}p_\lambda(x) = e^{-h_1}h_n$; however Stanley claims the RHS is $e^{-h_1}\sum_{n \geq 0}h_n$. I don't understand where this summation comes from. I'm sure it's something silly, but I've retraced my steps many times and can't find my mistake. Any help is greatly appreciated.