Is there any way to simplify the following fractions? Thank you!

Question

Is there any way to simplify the following fractions?

$[n^{3}-1]/[n^{4}-1]$

$n!/n^{n}$

Answer 1

1. Since $n^k-1 = 0$ when $n=1$, $(n-1)$ is clearly a factor. Then you can factor out $(n-1)$ by long division for example. This answer also gives a factorization of $n^k-1$.

2. Not much information is given except the common $n$.

Answer 2

For the first:

$$\frac{n^3-1}{n^4-1}=\frac{(n-1)(1+n+n^2)}{(n-1)(1+n+n^2+n^3)}=1-\frac{n^3}{1+n+n^2+n^3}$$

For the second: $$\frac{n!}{n^n}=\frac{(n-1)!}{n^{n-1}}=\frac{(1-\frac 1n)^{n-1}(n-2)!}{(n-1)^{n-2}}=\frac{(n-k-1)!\prod_{r=1}^{k}(1-\frac{1}{n-r+1})^{n-r}}{(n-k)^{n-k-1}}$$ for $k \in \Bbb N$