Prove that the derivative of $\frac{1}{z^n}$ at $z = a$ for any positive integer n and any non-zero complex number $a$ is $−na^{−n−1}$ using complex limits and the derivative formula for $z^n$.
I will use proof by induction for this problem. First, we will prove the base $n=1$. $$f(z)=\frac{1}{z}$$ should result in $f'(a)=-a^{-2}$. The complex definition of a derivative states: $$\lim_{h\to0}\frac{f(z+h)-f(z)}{h}.$$ So the complex derivative of our $f(z)$ should be $$\lim_{h\to0}\frac{\frac{1}{z+h}-\frac{1}{z}}{h}.$$ Finding a common denominator and subtracting the fractions in the numerator will result in: $$\lim_{h\to0}\frac{\frac{z}{z^2+zh}-\frac{z+h}{z^2+zh}}{h}=\lim_{h\to0}\frac{\frac{-h}{z^2+zh}}{h}.$$ Simlifying this fraction gives $$\lim_{h\to0}\frac{-h}{h(z^2+zh)}=\lim_{h\to0}\frac{-1}{z^2+zh}.$$ Finally, we can set $h=0$ and find that $$f'(z)=\lim_{h\to0}\frac{-1}{z^2+zh}=\frac{-1}{z^2}$$ and that $f'(a)=-a^{-2}$ so our base case is confirmed.
Now we will try to prove this to be true for the next term $n=k+1$. This gives $f(z)=\frac{1}{z^{k+1}}$. Following the limit definition of a derivative $$\lim_{h\to0}\frac{\frac{1}{(z+h)^{k+1}}-\frac{1}{z^{k+1}}}{h}.$$ Once again simplifying our numerator by subtracting the fractions gives $$\lim_{h\to0}\frac{\frac{z^{k+1}}{z^{k+1}(z+h)^{k+1}}-\frac{(z+h)^{k+1}}{z^{k+1}(z+h)^{k+1}}}{h}=\lim_{h\to0}\frac{\frac{z^{k+1}-(z+h)^{k+1}}{z^{k+1}(z+h)^{k+1}}}{h}=\lim_{h\to0}\frac{z^{k+1}-(z+h)^{k+1}}{hz^{k+1}(z+h)^{k+1}}.$$ Using binomial expansion for $(z+h)^{k+1}=\sum_{q=0}^{k+1}z^{k+1-q}h^{q}$ gives $$\lim_{h\to0}\frac{z^{k+1}-\sum_{q=0}^{k+1}z^{k+1-q}h^{q}}{hz^{k+1}(z+h)^{k+1}}.$$ Rewriting the expansion as $\sum_{q=0}^{k+1}z^{k+1-q}h^{q}=z^{k+1}+\sum_{q=1}^{k+1}z^{k+1-q}h^{q}$ allows the $z^{k+1}$ terms to cancel to give $$\lim_{h\to0}\frac{z^{k+1}-z^{k+1}-\sum_{q=1}^{k+1}z^{k+1-q}h^{q}}{hz^{k+1}(z+h)^{k+1}}=\lim_{h\to0}\frac{-\sum_{q=1}^{k+1}z^{k+1-q}h^{q}}{hz^{k+1}(z+h)^{k+1}}.$$ From this summation, each term in the numerator will have at least one $h$ term, which allows the $h$ in the denominator to cancel. This gives $$\lim_{h\to0}\frac{-\sum_{q=1}^{k+1}z^{k+1-q}h^{q-1}}{z^{k+1}(z+h)^{k+1}}.$$
Finally, shifting the summation once again to get $\sum_{q=1}^{k+1}z^{k+1-q}h^{q-1}=z^k+\sum_{q=2}^{k+1}z^{k+1-q}h^{q-1}$ gives $$\lim_{h\to0}\frac{-(z^k+\sum_{q=2}^{k+1}z^{k+1-q}h^{q-1})}{z^{k+1}(z+h)^{k+1}}.$$ At this point one can plug in $0$ for $h$ into the limit to evaluate it as $$\frac{-z^k}{z^{k+1}z^{k+1}}=\frac{-z^k}{z^{2k+2}}=\frac{-z^k}{z^{2(k+1)}},$$ but I think my proof seems to fall apart here. plugging back in $n=k+1$ gives $$\frac{-z^{n-1}}{z^{2n}}$$ This simplifies to $$\frac{-1}{z^nz}.$$ Which is not what I am trying to solve the limit as. Not sure where to go from here.
There is an error in your binomial expansion, if written correctly using the same proof leads to the correct expression: $$\frac{-n}{z z^n}$$
(And you don't actually use induction here)