Does there exist a matrix $P$ such that $P^n=M$ for a special matrix $M$? Consider the matrix
$$
M=\left(\begin{matrix}
0&0&0&1\\
0&0&0&0\\
0&0&0&0\\
0&0&0&0\\
\end{matrix}\right).
$$
Is there a matrix $P\in{\Bbb C}^{4\times 4}$ such that $P^n=M$ for some $n>1$?

One obvious fact is that if such $P$ exists, then $P$ must be nilpotent. However, I have no idea how to deal with this problem. Furthermore, what if $M$ is an arbitrary nilpotent matrix with index $k$?
 A: Purely by trial and error, let
$$P=\left(\begin{matrix}
0&1&1&1\\
0&0&0&1\\
0&0&0&0\\
0&0&0&0\\
\end{matrix}\right)$$ which squares to $M$.
The matrix
$$P=\left(\begin{matrix}
0&1&0&0\\
0&0&0&1\\
0&0&0&0\\
0&0&0&0\\
\end{matrix}\right)$$
and its anti-transpose are more simple and also work.
A: For 
$P=\left(\begin{matrix}
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
0&0&0&0\\
\end{matrix}\right)$
$P^3=M$.
Can we find a matrix $P$ such that $P^n=M$ for some $n>3$? Or for which $n$ does there exist $P$ s.t $P^n=M$?
A: Since it has been proposed to treat this Question as a duplicate of the present one, it should be noted that there is a negative Answer to the "new" issue raised in Yeyeye's Answer here.
As earlier observed, since $M$ is nilpotent, for $P^n = M$ for $n\gt 1$ will require that $P$ is nilpotent.  It follows that the minimal polynomial for $P$ must divide $x^d$ where $d$ is the degree of nilpotency (i.e. the least power such that $P^d=0$.
Now the characteristic polynomial of $P$ will have degree $4$ (because $P$ is $4\times 4$), and thus $d\le 4$.  So the minimal polynomial of $P$ has the form $x^k$ for $1\lt k \le d \le 4$.  In other words, $P^4=0$ given the above information.  So $P^3 = M \neq 0$ is the largest power $n=3$ that one can achieve.
