Vandermonde matrix unique solution to polynomial equation 
I want to verify if I am thinking of parts b) and c) correctly. 
To find its kernel, we must find the nullspace of the matrix equation 
$
\begin{bmatrix}
5^5 & 5^4 & 5^3 & 5^2 & 5 & 1 \\
4^5 & 4^4 & 4^3 & 4^2 & 4 & 1 \\
3^5 & 3^4 & 3^3 & 3^2 & 3 & 1 \\
2^5 & 2^4 & 2^3 & 2^2 & 2 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 1 
\end{bmatrix}
\begin{bmatrix} 
a_5 \\
a_4 \\
a_3 \\
a_2 \\
a_1 \\
a_1 
\end{bmatrix} 
= 
\begin{bmatrix} 
0 \\
0 \\ 
0 \\
0 \\
0 \\
0 \\
\end{bmatrix}
$
Due to the fact that this is a vandermonde matrix with distinct geometric ratios for its rows, this matrix has full rank and thus has a trivial nullspace I think? 
Then, to show part c that there is a unique polynomial such that $f(j) = a_j$, can we then just say that 
$
\begin{bmatrix}
5^5 & 5^4 & 5^3 & 5^2 & 5 & 1 \\
4^5 & 4^4 & 4^3 & 4^2 & 4 & 1 \\
3^5 & 3^4 & 3^3 & 3^2 & 3 & 1 \\
2^5 & 2^4 & 2^3 & 2^2 & 2 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 0 & 1 
\end{bmatrix}
\begin{bmatrix} 
a_5 \\
a_4 \\
a_3 \\
a_2 \\
a_1 \\
a_1 
\end{bmatrix} 
= 
\begin{bmatrix} 
a_5 \\
a_4 \\ 
a_3 \\
a_2 \\
a_1 \\
a_0 \\
\end{bmatrix} $
has a unique solution due to nontrivial nullspace? 

Well, you are right saying that this is a Vandermonde matrix and hence is invertible, however, this exercise is basically how one proves this fact. You can use the formula to compute such a determinant (which does not use this exercise), but I guess this is not the spirit of the assignment. 
How about saying instead that $T$ is invertible, since if $f\in V$ is such that $T(f)=0$, then $f$ has $6>5$ distinct roots, so that $f=0$.
For the rest of the exercise, your solution is fine.
