Any Nilpotent Matrix is not Diagonalizable

First note that the only eigenvalue of a nilpotent matrix $A$ is $0$. Now suppose that $A$ is diagonalizable and nonzero. Since $A$ is diagonalizable, $A$ is similar to a diagonal matrix with the eigenvalues of $A$ on the main diagonal, i.e., $A=PDP^{-1}$ where $D$ is the zero matrix so that $A$ is the zero matrix, a contradiction.

Not true (example: $0$ matrix). Though, the statement is true if $A \neq 0$.

If $A$ is diagonalizable, then there exists an invertible $P$ and a diagonal matrix $D$ such that $A = P^{-1} D P$. Then, $A^N = P^{-1} D^N P$, which gives $P^{-1} D^N P = 0$. Therefore $D^N = 0$ and so $(D_{ii})^N = 0$ for all $i$, hence $D_{ii} = 0$ for all $i$ and $D$ is the zero matrix. This gives $A = P^{-1} 0 P = 0$, which is contradictory to the assumption.

The claim is true for nonzero matrices. The minimal polynomial is of the form $X^n$ for $n>1$ so it has repeated roots. Then your matrix cannot be diagonalizable. Alternatively all the eigenvalues are zero, so the only diagonalizable nilpotent matrix is the zero matrix.