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: 
I would group the higher components together using difference of squares from 10^2.

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

$\approx 10^{11}×0.75×0.84×0.9×24$

$\approx 10^{11}×0.57×24$

$\approx 1.3×10^{12}$
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.