Question: I need to find the approximate value of *15!* . In answer sheet i've got following answers: 

A) *1 307 xxx xxx xxx*

B) *1 207 xxx xxx xxx*

C) *1 405 xxx xxx xxx*

where x are futher digits in these numbers.

Frankly speaking I've no idea how to approximate this. After some research I've found Stirling's approximation but still without calculator i'm not able to get this result. Can you give me any tips ?

Answer: 
It is a question of grouping the factors into chunks which multiply to form "nice" numbers which are close to numbers having many zeros.

$$
15! = \underbrace{7 \times 13 \times 11} \times \underbrace{7 \times 9 \times 8 \times 2} \times 1296 \times 1000 \\ = 1001 \times 1008 \times 1296 \times 1000
$$

Now, it is really easy : note that $1001 \times 1008 \geq 1000^2$, but not by much. This gives exactly the first four digits being greater than or equal to $1296$, but not by much, so the answer should be expected to be $1 307...$
I would group the higher components together using difference of squares from 10^2.

$15! = (15×5)×(14×6)×(13×7)×(12×8)×(11×9)×10×4×3×2$

$\approx 10^{11}×0.75×0.84×0.9×24$

$\approx 10^{11}×0.57×24$

$\approx 1.3×10^{12}$