Negative power

What does this number mean? Which order of 1, 2, 3 and 4 makes the highest value? Which makes the lowest?

Problem

Take a look at this expression. What does it mean?  $$\left(\left(-4^{-3}\right)^{-2}\right)^{-1}$$

Do you think it means

$\left(\left(\left(-4\right)^{-3}\right)^{-2}\right)^{-1}$ (Interpretation A)    or   $\left(\left(-\left(4^{-3}\right)\right)^{-2}\right)^{-1}$(Interpretation B)?

 

Check your thinking

By usual mathematical conventions for orders of operations, $\left(\left(-4^{-3}\right)^{-2}\right)^{-1}$ means $\quad\left(\left(-\left(4^{-3}\right)\right)^{-2}\right)^{-1}$ (Interpretation B). You can include the extra brackets if you find these helpful, but make sure you put them in the right place.

 
Does it make a difference how we interpret $\left(\left(-4^{-3}\right)^{-2}\right)^{-1}$ ?
 
Check your thinking

Both $\left(\left(\left(-4\right)^{-3}\right)^{-2}\right)^{-1}$ and $\left(\left(-\left(4^{-3}\right)\right)^{-2}\right)^{-1}$ have the same value, so no, in this case it doesn't make a difference. 

 
Do we still get the same value if we change the order of $1, 2$ and $3$ in the powers?
 
Check your thinking

If we swap the order of $-3$ and $-2$, then Interpretation A gives us 

$$\left(\left((-4)^{-2}\right)^{-3}\right)^{-1}=\left(\left(\dfrac{1}{(-4)^{2}}\right)^{-3}\right)^{-1}=\left(\dfrac{1}{16}\right)^{(-3)\times(-1)}=\left(\dfrac{1}{16}\right)^3=\dfrac{1}{16^3}$$

But Interpretation B gives us

$$\left(\left(-(4^{-2})\right)^{-3}\right)^{-1}=\left(\left(-\dfrac{1}{4^{2}}\right)^{-3}\right)^{-1}=\left(-\dfrac{1}{4^{2}}\right)^{(-3)\times(-1)}=\left(-\dfrac{1}{16}\right)^3=-\dfrac{1}{16^3}$$ 

So in this case, it does matter how we interpret $\left(\left(-4^{-2}\right)^{-3}\right)^{-1}$.

 
Now think about changing the order of $1,2,3$ and $4$ in $\left(\left(-4^{-3}\right)^{-2}\right)^{-1}$    
 
Which order of $1,2,3$ and $4$ makes the highest value?
 
Which order makes the lowest value?
 
How many different values can you get?