Why my proof that $\lim_{x\rightarrow\infty}\frac{x!}{x^x}=1$ is wrong

Question

It is well known that $$\lim_{x\rightarrow\infty}\frac{x!}{x^x}=0$$But according to the definition of the Stirling numbers of the first kind, the falling factorial could be written as $$(x)_n=\sum_{k=0}^ns(n,k)x^k$$ Where $(x)_n=x(x-1)(x-2)...(x-n+1)$. So shouldn't $(x)_x$ be $x!$ with a $x^x$ term by the falling factorial expansion, making the limit $1$?

Answer 1

$$\frac{x!}{x^x}={\frac{x(x-1)(x-2)(x-3)\cdots}{x\cdot x\cdot x\cdot x\cdots}}=\frac{x-1}{x}\cdot\frac{x-2}{x}\cdots\frac{2}{x}\cdot\frac{1}{x}$$

Where there are $x$ terms in the numerator and denominator.

Cancelling the first term in both, we see that the limit approaches $0$ as each term in the product approaches $1/x$ with $x\to\infty$

To see that it cannot be $1$, note that you have an infinite product of numbers between $0$ and $1$.