The problem statement:

Notice that $\displaystyle n_0,\cdots,n_m$ need NOT be pairwise coprime.Let $\displaystyle n_0,\cdots,n_m\in\mathbb{Z}$. Prove that there is a unique $\displaystyle f(x)\in\mathbb{Q}[x]$ of degree $\displaystyle \leq m$ such that $\displaystyle f(i)=n_i$ for all $\displaystyle i=0,\cdots,m$. [Hint: Use the Chinese remainder theorem.]

It seems to me that this problem is equivalent to showing that the linear system $\displaystyle Ax=b$, where

$\displaystyle A=\left(\begin{array}{ccc}0^0&\cdots&0^m\\\vdots&\ vdots&\vdots\\m^0&\cdots&m^m\end{array}\right)$, $\displaystyle x=\left(\begin{array}{c}a_0\\\vdots\\a_m\end{array }\right)$ and $\displaystyle b=\left(\begin{array}{c}n_0\\\vdots\\n_m\end{array }\right)$,

has exactly one solution, which in turn is equivalent to showing that $\displaystyle A$ is invertible.

However, the hint tells me to use the Chinese remainder theorem---and I don't see how that's relevant.

Any help would be much appreciated. Thanks!