Approximate value of $15!$

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.

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

$\approx 10^{11}×0.75×0.84×0.9×24$

$\approx 10^{11}×0.57×24$

$\approx 1.3×10^{12}$