Sum of partial factorials: number of digits for $\sum_{n=1}^{256}\frac{256!}{(256 - n)!}$

Question

According to Wolfram Alpha, $256! \approx 8.5 \times 10^{506}$, i.e. it has $507$ digits. My question is how can one evaluate such approximations for this sum below?:

$\displaystyle \sum_{n=1}^{256}\frac{256!}{(256 - n)!} = 256 + 256 \times 255 + 256 \times 255 \times 254 + \dots + 256! $

Answer 1

$$\sum_{n=1}^{256}\frac{256!}{n!}\approx256!\sum_{n=1}^\infty\frac1{n!}=(e-1)\cdot256!$$