Fundamental period of $x[kn]$ Let $x[n]$ be a periodic with fundamental period $N$ and $y[n] = x[kn]$ where $k \in \mathbb{N}$ and $k\ge2$. Is $y[n]$ periodic? What's the fundamental period of $y[n]$?

Here is my answer:
If $N = mk$ where $m \in \mathbb{N}$ then $y[n+m] = x[k(n + \frac{N}{k})] = x[kn+N] = x[kn] = y[n]$. So in this case the fundamental period is $\frac{N}{k}$. Otherwise simplify $\frac{N}{k}$ such that $\frac{N}{k} = \frac{m}{n}$ where $\gcd(m , n) = 1$. The fundamental period is $m$ since $y[n+m] = x[k(n+\frac{nN}{k}) ] = x[kn + nN] = x[kn] = y[n]$.
I don't know whether my solution is correct and rigorous enough. For instance how can we prove there is no smaller period?
You are correct. The following is what you have shown.
Let $x[n]$ be a periodic sequence with period N :
$$ x[n] ~=~ x[n + rN] ~~~~~,~~~\forall r,N \in Z^+ $$
Define $y[n] = x[M n]$, where $M$ is a positive integer; then
$$ y[n] = x[Mn] = x[Mn + rN] = x[M(n + \frac{rN}{M})] = y[n+K] $$
$y[n]$ will be periodic with $K$, if $\frac{r ~N}{M}$ is an integer. Hence, set $r$ to any value that makes $\frac{rN}{M}$ an integer.
If $N,M$ are already coprime, then setting $r=M$ makes $K=N$ as the period of $y[n]$. If they are not already coprime, then make them coprime $M_c,N_c$, and then set $r=M_c$ to make $K = N_c$ as the period of $y[n]$.
Make $N,M$ coprime by dividing them by their greatest common divisor $g = \text{gcd}(N,M)$ :
$$ N_c = N / g ~~~,~~~~ M_c = M / g $$
Then, for the general case, the period of $y[n]$ is:
$$ \boxed{ K = \frac{N}{ \text{gcd}(N,M) } } $$