Let $b_n=\dfrac{A_n}{n!}$; then your recurrence can be rewritten
$b_n=1+\frac1n\sum_{k=1}^{n-1}b_k\;,\tag{1}$
with $b_1=1$. Calculate a few values:
$\begin{align*} &b_1=1\\ &b_2=1+\frac12\cdot1=\frac32\\ &b_3=1+\frac13\left(1+\frac32\right)=\frac{11}6\\ &b_4=1+\frac14\left(1+\frac32+\frac{11}6\right)=\frac{25}{12} \end{align*}$
I recognize those as the first four harmonic numbers: $b_n=H_n=\sum_{k=1}^n\frac1k\;.$
And sure enough, the harmonic numbers do satisfy $(1)$:
$\begin{align*} 1+\frac1n\sum_{k=1}^{n-1}H_k&=1+\frac1n\sum_{k=1}^{n-1}\sum_{i=1}^k\frac1i\\ &=1+\frac1n\sum_{i=1}^{n-1}\sum_{k=i}^{n-1}\frac1i\\ &=1+\frac1n\sum_{i=1}^{n-1}\frac{n-i}i\\ &=1+\frac1n\sum_{i=1}^{n-1}\left(\frac{n}i-1\right)\\ &=1+\sum_{i=1}^{n-1}\frac1i-\frac{n-1}n\\ &=1+H_{n-1}-1+\frac1n\\ &=H_n\;. \end{align*}$
Thus, $A_n=n!H_n$.
Added: Here’s another approach to the problem. It’s not too hard to prove that the number of $\pi\in S_n$ with $k$ cycles is $\left[n\atop k\right]$, a Stirling number of the first kind. These have the generating function $\sum_{k\ge 0}\left[n\atop k\right]x^k=x^{\overline{n}}=x(x+1)(x+2)\dots(x+n-1)\;.$ Differentiating yields
$\begin{align*} \sum_{k\ge 1}k\left[n\atop k\right]x^{k-1}&=\frac{d}{dx}\Big(x(x+1)(x+2)\dots(x+n-1)\Big)\\ &=\sum_{k=0}^{n-1}\frac{\prod_{i=0}^{n-1}(x+i)}{x+k}\;, \end{align*}$
and evaluating at $x=1$ yields $\sum_{k\ge 1}k\left[n\atop k\right]=\sum_{k=0}^{n-1}\frac{n!}{k+1}=n!H_n\;.$