$V$ is a linear space. Need to compute $T^n$ Given: $V$ is a linear space $(\dim V = n)$ and there is a linear transformation $T: V \rightarrow V$ that $T^n = 0$ and $T^{n-1} \ne 0$ , also there's $u \in V$ that $T^{n-1}(u) \ne 0$
Prove that $B = (u,T(u),T^2(u),\dots,T^{n-1}(u))$ is a base for V.

Well, I tried saying that there are scalars $a_0,a_1,\dots,a_{n-1}$ and:
$a_0u + a_1T(u) + ... + a_{n-1}T^{n-1}(u) = 0$
, So if I prove that $a_0=a_1=a_{n-1}$ = 0 then I can say because there are n elements and also it's linearly independent. Thus, B is a base.
Well, I tried using the function $T^{n-1} ,T^{n-2},...,T^0$ like that:
[Explanation]:
First, I use the function $T^{n-1}$ on $a_0u + a_1T(u) + ... + a_{n-1}T^{n-1}(u) = 0$
which means it will look like this:  $a_0T^{n-1}u + a_1T^n(u) = 0$
(Because, all the other elements will have a dimension higher than $DimV = n$ so I erased them)
But, $T^n = 0$ Thus,  $a_0T^{n-1}u = 0$ , From here I say $T^{n-1} \ne 0$ So $a_0 = 0$.
Now I use $a_0 = 0$ and I use the function $T^{n-2}$ on $ a_1T(u) + ... + a_{n-1}T^{n-1}(u) = 0$ And so on.. Until I get all the scalars equal to zero. Then, I can say it is a base.
What do you guys think? Is my explanation correct?
[The answer]:
First, I use the function $T^{n-1}$ on $a_0u + a_1T(u) + ... + a_{n-1}T^{n-1}(u) = 0$ 

which means it will look like this:  $a_0T^{n-1}u + a_1T^n(u) = 0$ 

(Because, all the other elements will have a dimension higher than $DimV = n$ so I erased them)  

But, $T^n = 0$ Thus,  $a_0T^{n-1}u = 0$ , From here I say $T^{n-1} \ne 0$ So $a_0 = 0$.

Now I use $a_0 = 0$ and I use the function $T^{n-2}$ on $ a_1T(u) + ... + a_{n-1}T^{n-1}(u) = 0$ And so on.. Until I get all the scalars equal to zero. Then, I can say it is a base.
[The answer]:
First, I use the function $T^{n-1}$ on $a_0u + a_1T(u) + ... + a_{n-1}T^{n-1}(u) = 0$ 

which means it will look like this:  $a_0T^{n-1}u + a_1T^n(u) = 0$ 

(Because, all the other elements will have a dimension higher than $DimV = n$ so I erased them)  

But, $T^n = 0$ Thus,  $a_0T^{n-1}u = 0$ , From here I say $T^{n-1} \ne 0$ So $a_0 = 0$.

Now I use $a_0 = 0$ and I use the function $T^{n-2}$ on $ a_1T(u) + ... + a_{n-1}T^{n-1}(u) = 0$ And so on.. Until I get all the scalars equal to zero. Then, I can say it is a base.