Question 806080
Let's see if I follow you.
{{{ ( 1 / ( (2a)^-2 * b^2 * z^(-10) ) )^(-5) }}}
Hope I got it
A good way to look at this 
is that it is the same as
{{{ ((1 /  (2a)^-2) * (1 / b^2) * (1 / z^(-10) ))^(-5) }}}
Now I can use the fact that
{{{ ( a*b*c )^d = a^d * b^d * c^d }}}
{{{ (1 /  (2a)^-2)^(-5) * (1 / b^2)^(-5) * (1 / z^(-10) )^(-5) }}}
Now reverse the signs of the inner exponents and 
bring them up to the top
{{{ ((2a)^2)^(-5) * ( b^(-2))^(-5) * ( z^10 )^(-5) }}}
Now use the fact that
{{{ ( a^b )^c = a^(b*c) }}}
{{{ (2a)^(-10) * b^(10) * z^(-50) }}}
That's it