Suppose not. Then\[\forall i\in\{0, 1, \ldots, n + 1\}: -1 < a^i - P(i) < 1\implies -{n + 1 \choose i} < (-1)^i{n + 1 \choose i}(a^i - P(i)) < {n + 1 \choose i}.\]Sum from $i = 0$ to $n + 1$ and use the fact that the $n + 1$-th partial difference of a degree $n$ polynomial is zero to find\[-2^n < \sum_{i = 0}^{n + 1} (-1)^i{n + 1 \choose i}a^i < 2^n,\]so\[-2^n < (1 - a)^n < 2^n.\]Then clearly $|1 - a|^n < 2^n$ and $|1 - a| < 2$, contradicting $a \ge 3$! Similar things hold if $a \le -1$, actually.