= Solution
The <deletion projection for a symmetric group> removes $n$ from the cycle containing it. A cycle $(a_1\ \cdots\ a_r\ n)$ becomes $(a_1\ \cdots\ a_r)$, with a singleton understood as a fixed point; all other cycles remain unchanged. This describes a genuine <permutation> of $\{1,\ldots,n-1\}$ and immediately proves the two identity assertions.
Put $H=S_{n-1}$. For $\sigma,\theta\in H$ and $k<n$, if $\pi(\theta(k))\ne n$, deletion gives $\sigma\pi(\theta(k))$. If $\pi(\theta(k))=n$, it gives $(\sigma\pi\theta)(n)=\sigma\pi(n)$. These are exactly the two cases in $\sigma\pi_n\theta(k)$. Thus
$$
\boxed{(\sigma\pi\theta)_n=\sigma\pi_n\theta.}
$$
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