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 $1 307...$
In addition of what's said above, There is another way to approach nearly to less calculations:

$15!=3\*2\*(3\*2)\*7\*2^3\*3^2\*11\*(3\*2^2)\*13\*(7\*2)\*3\*10^3$

Occurrence of primes $2$ and $3$ necessarily is dense, this suggests we figure a way to route them all to one side. let's denote it $\delta=2^8\*3^6=12^4\*9$

$\delta/9=(5\*2+2)^4=10^4+2\binom{1}{4}10^3+2^2\binom{2}{4}10^2+2^3\binom{3}{4}10+2^4$=$10^4+2\*4\*10^3+4\*6\*10^2+8\*4\*10+16$
$=10^4+8\*10^3+24\*10^2+32\*10+16$

$15!=7\*7\*13\*(11\*9)\*10^3(delta/9)$=$49\*13\*(10^2-1)(10^7+8\*10^6+24\*10^5+32\*10^4+16\*10^3)$

You can see until this point, calculations are achieved only by hand.

$15!=(637\*100-637)\*polynomial=(630\*100+700-637)\*pol=(63000+63)\*pol$

You see now the trick is easy you may just evaluate the polynomial this way:

$1 0 0 0 0$

$0 8 0 0 0$

$0 2 4 0 0$

$0 0 3 2 0$

$0 0 0 1 6$

It sums to $20736$, multiplied to $63063$ then to 10^3 .It equals approximately $10^3(130.7\*10^7)$. This is the only point when you need extra-tools to multiply both factors.