I'm trying to solve the following problem:
Given $x_0,\ldots,x_n$ distinct real points. Choose a function $$P_n(x)=\sum_{j=0}^n c_j e^{jx}$$ s.t. $$P_n(x_i)=y_i \qquad i=0,\ldots,n $$ Show there's a unique choice of $c_0,\ldots,c_n$. Note that you can reduce this to polynomial interpolation.
To attack this, I was just thinking to write Taylor series for the exponential:
$e^{jx} = \sum_{k=0}^\infty \frac{(jx)^k}{k!}$
So the conditions become: $$\sum_j c_j \sum_{k=0}^\infty \frac{(jx_i)^k}{k!}=y_i$$ for $i=0,\ldots,n$. The issue I see is that the poly has an "infinite" degree, so I don't know how to move from here. Any hint is highly appreciated!
If $x_0, \cdots, x_n$ are distinct points, the Vandermonde matrix $$ \begin{pmatrix} 1 &x_0 & x_0^2& \cdots & x_0^n\\ 1 &x_1 & x_1^2 & \cdots &x_1^n\\ \vdots & \vdots& \vdots&\ddots& \vdots\\ 1& x_n & x_n^2 & \cdots & x_n^n \end{pmatrix} $$ is invertible. This is equivalent to say that the interpolating polynomial is unique. In your case, the set of conditions $P_n(x_i)=y_i$ leads to a linear system having the matrix
$$ \begin{pmatrix} 1 &e^{x_0} & e^{2x_0}& \cdots & e^{nx_0}\\ 1 &e^{x_1} & e^{2x_1} & \cdots &e^{nx_1}\\ \vdots & \vdots& \vdots&\ddots& \vdots\\ 1& e^{x_n} & e^{2x_n} & \cdots & e^{nx_n} \end{pmatrix}= \begin{pmatrix} 1 &e^{x_0} & (e^{x_0})^2& \cdots & (e^{x_0})^n\\ 1 &e^{x_1} & (e^{x_1})^2 & \cdots &(e^{x_1})^n\\ \vdots & \vdots& \vdots&\ddots& \vdots\\ 1& e^{x_n} & (e^{x_n})^2 & \cdots & (e^{x_n})^n \end{pmatrix} .$$This matrix is still invertible because $e^{x_0}, e^{x_1}, \cdots, e^{x_n}$ are distinct numbers.