2. Clearly $a_{1}>0$, and if $p \neq a_{1}$, we must have $a_{n}\left|a_{1}\right|$, and $p=-a_{n}$. But then for sufficiently large odd $k,-a_{n}^{k}=\left|a_{n}\right|^{k}>(n-1)\left|a_{1}\right|^{k}$, so that $a_{1}^{k}+\cdots+a_{n}^{k} \leq(n-1)\left|a_{1}\right|^{k}-\left|a_{n}\right|^{k}a_{1}$. From $a_{1}+\cdots+a_{n} \geq 0$ we deduce $\sum_{j=2}^{n}\left(x-a_{j}\right) \leq$ $(n-1)\left(x+\frac{a_{1}}{n-1}\right)$, so by the AM-GM inequality,
$$\left(x-a_{2}\right) \cdots\left(x-a_{n}\right) \leq\left(x+\frac{a_{1}}{n-1}\right)^{n-1} \leq x^{n-1}+x^{n-2} a_{1}+\cdots+a_{1}^{n-1}$$
The last inequality holds because $\binom{n-1}{r} \leq(n-1)^{r}$ for all $r \geq 0$. Multiplying (1) by $\left(x-a_{1}\right)$ yields the desired inequality.