We know that there are $n$ solutions for $n^{th}$ root of unity, $z^n = 1$ :
$1, z, z^2, z^3, ......z^{n-1}$ where, $z = e^{i2\pi/n}$
Now, $$z^{n+1} = z$$ right? Since, $$z^{n+1} = z^n.z = 1.z=z $$$$.:z^{n+1}=z$$
But if I write it this way, $$ z^{n+1} =(z^n)^{(n+1)/n}=(1)^{(n+1)/n}=1$$$$.:z^{n+1}=1$$
Which one is right? And so can I write,$$z^p = (z^n)^{p/n} = 1 ?$$where $p$ is any real number
There is no $n^{th}$-root function taking a complex number $z$ and returning $z^{1/n}$ such that $(z^n)^{1/n}=z$ for all $z\in \mathbb{C}$. This is because there are exactly $n$ $n^{th}$-roots (when $z\ne 0$). Taking an $n^{th}$-power of a complex number loses some information, namely you cannot tell which of the different $n^{th}$ roots of your result you had started with.
A simple example is when $n=2$ then $1^2=(-1)^2$ so that we can't decide between $+1$ and $-1$ for $1^{1/2}$.
