Let $E$ be a $\mathbb{C}$-vector space of dimension $n$. Let $g$ be an endomorphism of $E$ such that $g^n = 0$ and $g^{n-1} \neq 0$. Prove that there exists a basis of $E$ in which the matrix of $g$ is the matrix $N$ below: $$N = \left(\begin{array}{ccccc} 0 & 1 & 0 & \ldots & 0 \\ 0 & 0 & 1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ \vdots & & & \ddots & 1 \\ 0 & \ldots & & \ldots & 0 \end{array}\right)$$ In other words: $N = (n_{i,j})_{1 \leq i,j \leq n}$ with $n_{i,j} = 1$ if $j = i+1$ and $n_{i,j} = 0$ otherwise.
Let $E$ be a $\mathbb{C}$-vector space of dimension $n$. Let $g$ be an endomorphism of $E$ such that $g^n = 0$ and $g^{n-1} \neq 0$.
Prove that there exists a basis of $E$ in which the matrix of $g$ is the matrix $N$ below:
$$N = \left(\begin{array}{ccccc} 0 & 1 & 0 & \ldots & 0 \\ 0 & 0 & 1 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 0 \\ \vdots & & & \ddots & 1 \\ 0 & \ldots & & \ldots & 0 \end{array}\right)$$
In other words: $N = (n_{i,j})_{1 \leq i,j \leq n}$ with $n_{i,j} = 1$ if $j = i+1$ and $n_{i,j} = 0$ otherwise.