Expect a factorial

Lagrangian interpolation tells us that $\begin{aligned} f_n(x) & = \; \sum_{j=0}^n j! \prod_{{0 \le k \le n} \atop {k \neq j}} \frac{x-k}{j-k} \; = \; \sum_{j=0}^n \frac{j!}{(-1)^{n-j} j!(n-j)!}\prod_{{0 \le k \le n} \atop {k \neq j}} (x-k) =\sum_{j=0}^n \frac{(-1)^{n-j}}{(n-j)!} \prod_{{0 \le k \le n} \atop {k \neq j}}(x-k) \end{aligned}$ By the Taylor series for $e^x$ , the leading coefficient, $a_n$ of $f_n(x)$ approaches $\frac 1e$ for large $n$ . $f_{n}(x) - f_{n-1}(x)$ is a degree $n$ polynomial with roots at $x=0,1,2,...,n-1$ , so $f_{n}(x) - f_{n-1}(x) = a_nx(x-1)(x-2)\cdots (x-(n-1))$ . Setting $x=n$ gives us the numerator of the desired limit, so we have $\lim_{n\to\infty} \dfrac{n! - n!^e}{n!}=\lim_{n\to\infty} \dfrac{ a_nn!}{n!}=\frac 1e$

Incidentally, $n! - n!^e$ happens to be equal to the derangement function , $!n$ .