Number of real roots

For $n \in \{1,2,\ldots, 999\},$ $\begin{aligned} P_{999}(-n) &= 1 + (-1)^1\frac{n}{1!} + (-1)^2\frac{n(n-1)}{2!} + \cdots + (-1)^{999}\frac{n(n-1)\cdots(n-998)}{999!} \\ &= \binom{n}{0}(-1)^0 + \binom{n}{1}(-1)^1 + \binom{n}{2}(-1)^2 + \cdots + \binom{n}{999}(-1)^{999} \\ &= \sum_{k=0}^{999} \binom{n}{k} (-1)^k \\ &= \sum_{k=0}^n \binom{n}{k} (-1)^k + \sum_{k=n+1}^{999} 0\cdot (-1)^k\\ &= \big(1+(-1)\big)^n + 0\\ &= 0 \end{aligned}$

Therefore each of $-1, -2, \ldots, -999$ is a real root of $P_{999},$ which is a polynomial of degree $999,$ so there are exactly $999$ real roots of $P_{999}(x)$