How prove this matrix $A$ similar with $A^T$ Show that :
Matrix $A$ Similar $A^T$.
where
$$A=\begin{bmatrix}
0&1&0&\cdots&0\\
0&0&\cdots&\cdots&0\\
0&0&\cdots&1&0\\
\cdots&\cdots&\cdots&\cdots&1\\
0&0&0&\cdots&0
\end{bmatrix}_{n\times n}$$
It is clear the eigenvalue all 0 with the matrix $A$.maybe we can consider the Jordan block?
 A: This matrix $A$ sends the standard basis $e_1,e_2,...e_n$ to $Ae_1=0, Ae_2=e_1,...,Ae_n=e_{n-1}$.
The matrix $A^T$ sends them to $A^T e_1=e_2, A^T e_2=e_3,..., A^T e_n=0$.
The similarity consists in finding a matrix $P$ such that $A^T=P^{-1}AP$, i.e. $PA^T=AP$.
We can use this and the equations above to compute the columns of $P$. Notice the columns of $P$ are $Pe_1,Pe_2,... Pe_n$.
For example, we take the equation $PA^T=AP$ and apply it to $e_n$. We get
$$0=P0=PA^Te_n=APe_n.$$
So, we can put $Pe_n=e_1$.
Let us apply it now to $e_{n-1}$. We get
$$e_1=Pe_{n}=PA^Te_{n-1}=APe_{n-1}$$
So, we can put $Pe_{n-1}=e_2$.
Continue with $e_{n-2},...,e_1$ and you get all columns of $P$.
You can check that 
$$P=\begin{bmatrix}0&...&0&1\\0&...&1&0\\.&...&.&.\\1&...&0&0\end{bmatrix}$$

There is a way to make fix Erick Wong's comment so that it leads to a solution. Let us change it to 

What is the smallest $k$ such that $(A^T)^k=0$?

The answer is the same that you gave for the case of $A^k$. From this is follows that the Jordan canonical form of $A^T$ has a single block of size $n$ associated to the eigenvalue $0$, i.e. the Jordan canonical form of $A^T$ is $A$. In particular they are similar.
