I am thinking about the definition of a nilpotent matrix, what I know is that a nilpotent matrix is a square matrix, say matrix $A$, such that $A^k$ is a null matrix with index $k$.
According to this, there is no restriction about $A$ to be square matrix whose entries are all zeros, so $A$ can be nilpotent with index $0$. Can it be?
Say we have $A$, which is a $5 \times 5$ (or any other dimension)? Shall we find $A^2,A^3,...$ until we find the null matrix? I could not find any short-cut, is there any?
I think you meant, "index $k=$1". By convention $A^0=\mathrm{Id}$ so no matrix is nilpotent with index zero. If $A$ is an $n\times n$ square matrix, then $A$ has $n$ eigenvalues (up to multiplicity). It is necessary and sufficient that all the eigenvalues are zero for $A$ to be nilpotent (of any index).
The nilpotency index will be the maximum of the nilpotency indices (=size, in this context) of each Jordan block of $A$. To my knowledge, there is no further shortcut to deduce the size of the blocks aside from actually computing the Jordan block. Yes, you might compute the minimal polynomial... but if you can do that, you already know the nilpotency index!