Proving by induction that strictly lower triangular matrix is nilpotent 
Let matrix $A \in \mathbb{K}^{n \times n}$ have the property $a_{ij} = 0$ for $1 \leq i \leq j \leq n$. Show that $A^n = 0$.

Proof by induction:
Base Case: 
for $n=2: 
   A=
  \left[ {\begin{array}{cc}
   0 & 0 \\
   a_{21} & 0 \\
  \end{array} } \right]$
So $A^2 = \left[ {\begin{array}{cc}
   0 & 0 \\
   a_{21} & 0 \\
  \end{array} } \right] \cdot 
\left[ {\begin{array}{cc}
   0 & 0 \\
   a_{21} & 0 \\
  \end{array} } \right] = 
\left[ {\begin{array}{cc}
   0 & 0 \\
   0 & 0 \\
  \end{array} } \right]$
Inductive Hypothesis(IH):
Assume $A^n = 0$ holds true for some $n$.
Inductive Step:
$n \rightarrow  n+1$, to show: $A^{n+1} = 0 $ 
$A^{n+1} = A^n \cdot A =^{IH} 0 \cdot A = 0$
It seems to be too simple. Is it correct to prove this by induction?
 A: Yes, induction is certainly a good way to try and prove the result. However, there are some problems with your attempt to do so.
One problem is that the $\ A\ $ in your induction hypothesis is an $\ n\times n\ $ matrix, whereas the one in your induction step has to be an arbitrary $\ (n+1)\times(n+1)\ $ matrix satisfying the stated conditions. That means it cannot be the same $\ A\ $ as the one appearing in the induction hypothesis.
To fix the proof, your induction hypothesis must be something like $"\ A^n = 0\ $ for all strictly lower triangular $\ n\times n\ $ matrices $\ A\ $."  Also, in the induction step, an identity you will probably find useful is
$$
\pmatrix{B&0 _{n\times1}\\b^\top&0}^k=\pmatrix{B^k&0 _{n\times1}\\b^\top B^{k-1}
&0}\ ,
$$
for any $\ n\times n\ $ matrix $\ B\ $ and $\ n\times1\   $ column vector $\ b\ $.
A: If you try to deduce the $n=3$ case from $n=2$ that you describe, this happens:
$$
\begin{pmatrix}
0 & 0 & 0 \\
a_{21} & 0 & 0 \\
a_{31} & a_{32} & 0
\end{pmatrix}^{3} = 
\begin{pmatrix}
0 & 0 & 0 \\
a_{21} & 0 & 0 \\
a_{31} & a_{32} & 0
\end{pmatrix}^{2}
\begin{pmatrix}
0 & 0 & 0 \\
a_{21} & 0 & 0 \\
a_{31} & a_{32} & 0
\end{pmatrix}
$$
You can't conclude that $A^2 = 0$ because this matrix is $3\times 3$ and not $2\times 2$ like your previous case. 
