Proof that all $n\times n$ matrices that are nilpotent of order $n$ are similar. Can someone give a proof that all $n\times n$ matrices that are nilpotent of order $n$  are similar? 
A matrix $A$ is called nilpotent if there exists some positive integer $k$ such that $A^k$ is the $0$-matrix. The order of nilpotency of a matrix is the smallest $k$ with this property. That is, the question is about matrices such that $A^n=0$ yet $A^{n-1} \neq 0$. 
I am new in linear algebra and know up to linear transformation and isomorphisms and matrix representation of transformations. The same question is answered in this site using Jordan Matrix and other things which I do not know. So can anyone give a proof using elementary things which I know.
 A: If $A$ is an $n\times n$ matrix that is nilpotent of order $n$, then there is a non-zero vector $x$ such that $A^nx =0$ but $A^{n-1}x \ne 0$. So $\{ x,Ax,A^2x,\cdots,A^{n-1}x\}$ is a basis, and the representation of $A$ with respect to this basis has a standard form. This standard form is the same for any such matrix, which makes any two such matrices similar.
A: First a lemma: Consider the chain
$$
\ker(I)\subseteq \ker(A)\subseteq\ker(A^2)\subseteq \cdots\subseteq\ker(A^n)
$$
of subspaces of $\Bbb R^n$ (or $\Bbb C^n$ if these are complex matrices). Only the last kernel is the whole space, and the first kernel is the zero space. I claim that each inclusion is strict, and as a consequence the dimension has to increase by exactly $1$ each step.
Proof of claim: Assume that for some $k$ we have $\ker(A^k)=\ker(A^{k+1})$. Let $v\in\ker(A^{k+1})$. Then $Av\in\ker(A^{k+1})=\ker(A^k)$, which is to say $A^k(Ak)=0$, so $v\in\ker(A^{k+1})$. This shows that $\ker(A^{k+1})=\ker(A^{k+2})$, and by induction we see that in the chain above, from $\ker(A^k)$ on, there is only equality. Since $\ker(A^{n-1})\neq\ker(A^n)$, equality can't happen before then.
Corollary: $\ker(A^k)=\operatorname{im}(A^{n-k})$. And in particular, $A$ maps $\ker(A^{k})$ surjectivity onto $\ker(A^{k-1})$.
Now for the actual proof: Let $v_1$ be a non-zero element in the kernel of $A$, and let $v_k$ for $1< k\leq n$ be such that $Av_k=v_{k-1}$. In other words, let $v_2$ be such that $Av_2=v_1$, and so on (by the corollary above, such a vector can always be found). Finally, let $V$ be the matrix where the $k$'th column is $v_{k}$ (so that $v_1$ is the rightmost column). Then
$$
V^{-1}AV=\begin{bmatrix}
0&0&0&0&\cdots&0&0\\
1&0&0&0&\cdots&0&0\\
0&1&0&0&\cdots&0&0\\
0&0&1&0&\cdots&0&0\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&0&\cdots&0&0\\
0&0&0&0&\cdots&1&0\end{bmatrix}
$$
