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?

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. 
