# Does this prove that the factorial grows faster than the exponential?

I want to prove that the factorial grows faster than the exponential function. First, I introduce the ratio $$L = \frac{n!}{e^n}.$$ Then, I introduce another ratio : $$\frac{(n+1)!}{e^{n+1}} = \frac{(n+1) \cdot n!}{e \cdot e^n} = \frac{(n+1)}{e} L.$$ When the value of $$n$$ take values bigger and bigger and $$L$$ gets bigger and bigger. In other words, $$\lim_{n \to \infty} \frac{(n+1)}{e} = \infty,$$ meaning that $$L$$ is divergent. Thus $$n!$$ grows faster than the exponential.

• Yes, your idea is exactly right, but you need to give $L$ a better name, which depends on $n$ (like $L_n$). Nov 27, 2023 at 23:34

$$\frac{n!}{e^n} = \prod_{i=1}^n\frac{i}{e}.$$ Since $$\lim_{i \to \infty} \frac{i}{e} = \infty,$$ clearly the factorial is growing faster than the exponential.
Yes, your method shows specifically that $$n!=\omega(e^n)$$, since $$L_n\to\infty$$. There is nothing special about $$e$$ here, so you can use the same idea to show that $$n!=\omega(\alpha^n)$$ for any $$\alpha$$ (or to put it another way, $$n!=\omega(e^{\beta n})$$ for any $$\beta$$).