In honour of 2187

$\log_3 2187$ can be written as :

$\log_3 3^7$

this is equal to:

$7$

putting $7$ in place of $\log_3 2187$ ,

$(7-0)(7-1)(7-2)....(7-7)....(7-2186)$

since $7-7 = 0,$

the whole expression becomes $\boxed{zero}$ .(required answer)

In The problem $(\log_3 2187-0)(\log_3 2187-1)\cdot \cdot \cdot (\log_3 2187-2185)(\log_3 2187-2186)$ There may be one bracket of $(\log_3 2187-7)$ $=(\log_3 2187-\log_3 3^{7})$ $=(\log_3 2187-\log_3 2187)$ $=0$ Therefore by zero product property the value of the expression is $\boxed{=0}$

log(2187-2186) = log1 = O