Binomial Expansion on $(\frac{\rho}{r})^p$+$(\frac{z}{h})^p$ $=1$

Question

Good day all!

I have a quick question about this problem I ran into:

Think of $z$ as a function of $ \rho $. Solve the following equation for z and use binomial expansion to determine how $ z $ depends on $ \rho $ up to leading order.

Im given the formula :$(\frac{\rho}{r})^p+(\frac{z}{h})^p=1$.

So first I solve the formula for $z ( \rho)$ and use binomial expansion:

$ z( \rho)=h(1-( \frac{\rho}{r})^p)^{\frac{1}{p}}$

I am told to use the formula for binomial expansion: $(1+z)^n=1+nz+ \frac{n(n-1)}{2!}z^2+...$

What is meant by "determine how $ z $ depends on $ \rho $ up to leading order" What is leading order? Also, do I just use the formula as is and let $z=(\frac{\rho}{r})^p$ ?

Thanks for any help in advance!