<s>[INST] Expansion of $(1-z)^{-m}$ 
Expand $(1-z)^{-m}$, $m$ a positive integer, in powers of $z$.

Since $\dfrac{1}{1-z}=1+z+z^2+\ldots$, we can find
$$\dfrac{1}{(1-z)^2} = (1+z+z^2+\ldots)(1+z+z^2+\ldots) = 1+2z+3z^2+\ldots.$$
So, in a similar way, by using induction and multiplying by $1+z+z^2+\ldots$ at each step, we can find
$$\dfrac{1}{(1-z)^n} = 1+\binom{n}{n-1}z+\binom{n+1}{n-1}z^2+\binom{n+2}{n-1}z^3+\ldots.$$
I wonder whether this is a rigorous solution to the question or not. If not, what needs to be changed?
 [/INST] As written this isn't really rigorous. You hint at using induction to prove this, so you should actually demonstrate (by writing down an induction argument) that it works, instead of just showing it for a few cases.
That is, suppose that
$$\frac{1}{(1-z)^n} = 1 + \binom{n}{n-1}z + \binom{n+1}{n-1}z^2 + \binom{n+2}{n-1}z^3 + \cdots$$
and then derive from this that
$$\frac{1}{(1-z)^{n+1}} = 1 + \binom{n+1}{n}z + \binom{n+2}{n}z^2 + \binom{n+3}{n}z^3 + \cdots$$
You should also stipulate that this is only valid when $\left| z \right| < 1$.
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