Powers of lower triangular matrices How can I show that for an $n\times n$ strictly lower triangular matrix $A$, $A^n = [0]$, but $A^{n-1} \neq 0$? I can see it from some quick examples but  I'm having trouble formalizing those observations. 
 A: The $3 \times 3$ matrix
$$\left[\begin{array}{cc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0\end{array}\right]$$
gives a counterexample to the claim.
A: See Corollary 3.78 in my Notes on linear algebra for the proof of $A^n = 0_{n \times n}$. (The proof is not difficult, but is significantly lengthened by the fact that I could assume nothing for granted in that class.)
As @T.Bongers pointed out in his answer, $A^{n-1} \neq 0_{n \times n}$ is not always satisfied. However, for each $n > 0$, there exists a lower-triangular $n \times n$-matrix $A$ that satisfies $A^{n-1} \neq 0_{n \times n}$. To prove this, set
\begin{equation}
A = \begin{pmatrix}
0 & 0 & 0 & \cdots & 0 & 0 \\
1 & 0 & 0 & \cdots & 0 & 0 \\
0 & 1 & 0 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \cdots & 0 & 0 \\
0 & 0 & 0 & \cdots & 1 & 0
\end{pmatrix} ;
\end{equation}
this is the $n \times n$-matrix whose $\left(i,j\right)$-th entry is $\delta_{i, j+1}$ for all $i$ and $j$ (where we are using the Kronecker delta notation; see Definition 2.37 in op. cit.). Then, the $\left(i,j\right)$-th entry of $A^k$ is $\delta_{i, j+k}$ for all $i$ and $j$ and $k \geq 0$ (indeed, this can be proven by induction on $k$, where the induction step relies on the definition of the product of two matrices). Thus, in particular, the $\left(n,1\right)$-th entry of $A^{n-1}$ is $\delta_{n, 1+\left(n-1\right)} = \delta_{n, n} = 1 \neq 0$; therefore, $A^{n-1} \neq 0_{n\times n}$. (See also Show that $A^n = 0$ but $A^{n-1} \neq 0$ for an $n\times n$ strictly lower triangular matrix for variants of this proof.)
