Prove by induction that $(z^n)^*=(z^*)^n$ for all positive integers of n.
My knowledge of proving things by induction is still growing, so I wasn't really too sure on how to tackle the question as was quite different o the ones I've seen before.
Any help would be grateful.
For induction you need the show that, for the smallest value of $n$ allotted, that the equality holds. So for $n=1$,
$$(z^1)^*=z^*=(z^*)^1$$.
Now, for the induction phase you make the assumption that for some positive integer $k$, $(z^k)^*=(z^*)^k$. The final step is to compute $(z^{k+1})^*$ using the assumption provided...
$$(z^{k+1})^*=(z^k\cdot z)^*=(z^k)^*z^*=(z^*)^kz^*=\underbrace{z^*\cdot z^*\cdots z^*}_{k+1\text{ times}}\cdot z^*=(z^*)^{k+1}$$
where the second to last step is $z^*$ multiplied $k+1$ times. We also used the fact (unproven) that for two complex numbers $z,w$,
$$(zw)^*=z^*w^*$$
Thus the proof is complete.