Irish Mathematical Olympiad

Since
$$\lim_{x \to \infty} \frac{p(x)}{x^n} = a_n > 0,$$
$p(x)$ is positive for all sufficiently large positive $x$. But, in any event, $p(0) = -1$. Hence, by the Intermediate Value theorem, $p$ has at least one positive root. Suppose $a, b$ are two positive roots of $p$, so that $p(a) = p(b) = 0$. Then
$$0 = p(a) - p(b) = \sum_{i=0}^{n} a_i (a^i - b^i) = (a-b) \sum_{i=0}^{n} a_i \sum_{k=0}^{i-1} a^{i-k-1} b^k = (a-b)r(a,b),$$
where $r(a, b)$ is a positive expression. Hence, $a = b$, which means that $p$ has at most one positive root. The claimed result follows.