Proving linearly independence of the functions $t^{i}e^{\lambda_{0}t}$ Following the question I asked here and is:

Let $P(\lambda)=(\lambda-\lambda_{0})^{r}$where $r$ is a positive
  integer. Prove that the equation $P(\frac{d}{dt})x(t)=0$ has solutions
  $t^{i}e^{\lambda_{0}t},i=0,1,\ldots,r-1$

I now wish to prove the solutions are  linearly independent.
I have two questions regarding this: 


*

*I learned to prove such independence with the Wronskian, but I am having trouble calculating it in (I calculated the derivatives of $e^{\lambda_{0}t},te^{\lambda_{0}t}$ but its getting too hard when it is a greater power of $t$ since I am getting longer and longer expressions). How can I calculate the Wronskian ?

*If I think of the vector space that is the smooth real valued functions then it seems that this set (if I take the power of $t$ to be as big as I want, but finite) is linearly independent. did I deduce right ?
I would appriciate any help!
 A: To show that $t^i e^{\lambda_0t}$ are linearly independent, it suffices to show that $t^i$ are linearly independent.
A: Since you have some doubts, I'll try to give you a longer answers and maybe clear them.
The assertion that the $n$ functions $f_k(t)=t^k e^{\lambda_0 t} \text{ ; }{k=0,1,\dots,n-1}$ are linearly independent is that if
$$\sum_{k=0}^{n-1} c_k f_k(t)= e^{\lambda_0 t}(c_0+c_1 t +c_2 t^2+\cdots+c_{n-1} t^{n-1})=0$$
then
$$c_0=c_1=\cdots=c_{n-1}=0$$
Since $e^{\lambda_0 t}\neq 0$ for any $t$, it suffices to prove that if
$$c_0+c_1 t +c_2 t^2+\cdots+c_{n-1} t^{n-1}=0$$
then $$c_0=c_1=\cdots=c_{n-1}=0$$ or that the Wronskian determinant of the $n$ functions, $p_k(t)=t^k \text{ ; }{k=0,1,\dots,n-1}$ is never zero $0$.
For example, for the case $n=3$, we have the functions
$$y_0(t)=c_0$$
$$y_1(t)=c_1 t$$
$$y_2(t)=c_2 t^2$$
The Wronskian determinant is 
$$W(y_1,y_3,y_3)=\begin{vmatrix} {1}& {  t}  &{  t^2}  \\ {0}& {1}  &{2  t}  \\ {0}& {0}  &{2 } \end{vmatrix}=2 \cdot 1 \cdot 1 = 2! 1!$$
since all other combinations will occurr with a $0$. 
You can try and prove the Wronskian determinant
$$\begin{vmatrix}
   1 & t &  \cdots  & {{t^{n - 2}}} & {{t^{n - 1}}}  \\ 
   0 & 1 &  \cdots  & {\left( {n - 2} \right){t^{n - 3}}} & {\left( {n - 1} \right)}  \\ 
    \vdots  &  \vdots  &  \vdots  &  \vdots  &  \vdots   \\ 
   0 & 0 & 0 & {\left( {n - 2} \right)!} & {\left( {n - 1} \right)!t}  \\ 
   0 & 0 & 0 & 0 & {\left( {n - 1} \right)!}    \end{vmatrix}$$
will be equal to $1! 2! 3! \cdots (n-1)!$ so it cannot be zero.
Alternatively, one can prove the result by induction. You can assume the result is proven for $1,2,\dots, n-2$ and show the result is true for $n-1$. I think it is much easier to use the Wronskian determinant.
