If $A$ is an upper triangular matrix with diagonal entries are equal to $0$, why $A^n$ is equal to $0$ $${\bf A}_{n\times n} = \underbrace{
                \left.\left( 
                      \begin{array}{ccccc}
                             0&A_{1,2}&A_{1,3}&\cdots &A_{1,n}\\
                             0&0&A_{2,3}&\cdots &A_{2,n}\\
                             0&0&0&\cdots &A_{3,n}\\
                             \vdots&\vdots&\vdots&\ddots&\vdots\\
                             0&0&0&\cdots &0
                      \end{array}
                \right)\right\}
              }_{n\text{ columns}} 
              \,n\text{ rows}
$$
Let $A$ be an $n\times n$ upper triangular matrix and whose diagonal entries are $0$, 
then how could we prove that $A^n = 0$?
 A: The eigenvalues of an upper triangular matrix are the diagonal entries. Thus, the characteristic equation of $A$ is $\lambda^n$. According to the Cayley-Hamilton theorem, this implies that $A^n = 0$.
A: Another direction: as you may know, the $k$-th column of a matrix is the image of $(e_k)$, the $k$-th vector of the canonical basis of $\mathbb{R}^n$ (having all its coordinates zero, but the $k$-th which is 1).
Let $E_k$ be the subspace of $\mathbb{R}^n$ that contains all vectors having their $k$ last components equal to 0. 
Thus, Matrix A sends $V_1,V_2,... V_n$ to vectors that are all in $E_1$.
This is the first level. Now, if we reapply $A$ to any vector or $E_1$,
$$\left(  \begin{array}{ccccc}
                             0&A_{1,2}&A_{1,3}&\cdots &A_{1,n}\\
                             0&0&A_{2,3}&\cdots &A_{2,n}\\
                             \vdots&\vdots&\vdots&\ddots&\vdots\\
                             0&0&0&\cdots &A_{n-1,n}\\
                             0&0&0&\cdots &0
                      \end{array}
                \right)
\left(  \begin{array}{ccccc}
                             x_1\\
                             x_2\\
                             \vdots\\
                             x_{n-1}\\
                             0
                      \end{array}
                \right)$$
do you see why we obtain a vector of $E_2$, i.e., with its two last coordinates zero ?
By a very simple recurrence reasoning, we can establish that there is indeed a progressive dimension shrinking, one at a time, with $E_1 \supset E_2 \supset E_3 \cdots \supset E_k \cdots$.At the end, we are left with vectors having all their components equal to zero... 
A: The first basis vector is mapped to the zero vector; its image under the linear transformation is $0$.
The second basis vector is mapped to a scalar multiple of the first basis vector, and the first basis vector is mapped to $0$.  The image of the image of the second basis vector is $0$.
The third basis vector is mapped to a linear combination of the first two basis vectors.  Both of those are mapped to $0$ in two steps, so the third one is mapped to $0$ in three steps.
Continue in that way.
PS: If you prefer a slightly different rhythm of thought, you can put it like this: Call the standard basis vectors $e_1,\ldots,e_n$.  Then we have
$$
A e_1 = 0,
$$
$$
A e_2 = s e_1 \text{ (where $s$ is some scalar), so } A^2 e_2 = A(se_1) = 0.
$$
$$
A e_3 = s e_1 + s e_2 \text{ (where $s$ and $s$ are two different scalars), so } A^3 e_3 = A^2 (se_1+se_2) = 0.
$$
$$
A e_4 = (se_1+se_2+se_3) \text{ (where $s$ and $s$ and $s$ are three different scalars, so}
$$
$$
A^4 e_4 = A^3(se_1+se_2+se_3) = 0
$$
and so on.
A: What is $(A^2)_{1,2}$? What about $(A^2)_{2,3}$? Can you see what is happening here? How would this apply to $A^3$?
A: Denote by $A^{(r)}_{i,j}$ the coefficient at place $(i,j)$ in $A^r$. Then, for $i<n$,
\begin{align}
A^{(2)}_{i,i+1}
&=\sum_{1\le k\le n} A_{i,k}A_{k,i+1}\\[4px]
&=\sum_{1\le k<i} A_{i,k}A_{k,i+1}+
A_{i,i}A_{i,i+1}+A_{i,i+1}A_{i+1,i+1}+
\sum_{i+1<k\le n} A_{i,k}A_{k,i+1}
\end{align}
which is $0$ because for $1\le k<i$ we have $A_{i,k}=0$, $A_{i,i}=A_{i+1,i+1}$ and for $i+1<k\le n$ we have $A_{k,i+1}=0$.
Now suppose by inductive hypothesis that all coefficients $A^{(r)}_{i,j}=0$, for $i\le j\le i+r-1\le n$. Then, for $i\le j\le i+r$,
\begin{align}
A^{(r+1)}_{i,j}
&=\sum_{1\le k\le n}A^{(r)}_{i,k}A_{k,j}\\[4px]
&=\sum_{1\le k<i}A^{(r)}_{i,k}A_{k,j}+
\sum_{i\le k\le j}A^{(r)}_{i,k}A_{k,j}+
\sum_{j<k\le n}A^{(r)}_{i,k}A_{k,j}
\end{align}
