Matrix with all eigenvalues equal to $0$ is nilpotent. Let $A \in \mathbb{R}^{n \times n}$ be a matrix with all eigenvalues equal to zero. How does it follows that $A$ is nilpotent? In my text this is used without further argument but I don't see this straight away. 
 A: If a matrix $A\in\Bbb{R}^{n\times n}$ has the eigenvalue $0$ with multiplicity $n$ then it is nilpotent. After all, this means that $0$ has algebraic multiplicity $n$, so it is an $n$-fold root of the characteristic polynomial $P_A(X)$ of $A$. This is a polynomial of of degree $n$ so $P_A(X)=X^n$. Cayley-Hamilton tells us that $A^n=P_A(A)=0$, so indeed $A$ is nilpotent.
Note that the phrase "all eigenvalues equal to zero" is a bit ambiguous. The real matrix
$$\begin{pmatrix}0&1&0\\-1&0&0\\0&0&0\end{pmatrix},$$
has all real eigenvalues equal to zero, but it is not nilpotent.
A: A very explicit way to do so would use the Jordan normal form $J$ of your matrix, which looks like 
$$ \left( \begin{array}{ccccc} 0 & 1 & 0 & ...
\\ 0 & 0  & 1  & 0 & ...
\\ \vdots & & \ddots & \ddots
\\ 0 & ... & & 0 \end{array} \right).$$
There may be several 1's on the off-diagonal, or also zeroes. If you multiply this matrix with itself, the row of 1's moves one step to the upper right. So you can convince yourself that the $n$-th power of this matrix is the zero matrix. Since we have $A^n = T J^n T^*$ with the transformation matrices $T$, it follows also that $A^n = 0$.
