It does! And you can even make the stronger statement that the resulting polynomial is degree at most $n$ (just like with quadratics).
If you've seen systems of linear equations, there's an easy way to see why this is the case. Suppose you want a polynomial $f(x) = a_0 + a_1 x + \ldots + a_n x^n$ such that $f(x_0) = y_0$, $f(x_1) = y_1$, \ldots, $f(x_n) = y_n$ where we assume $x_i \neq x_j$ if $i \neq j$. The equations $f(x_i) = y_i$ are really $n+1$ equations in $n+1$ unknown variables $a_0,\ldots,a_n$ (i.e., in the coefficients of the unknown polynomial $f(x)$):
\begin{align}
a_0 + x_0 a_1 + \ldots + x_0^n a_n &= y_0 \\
&\vdots\\
a_0 + x_n a_1 + \ldots + x_n^n a_n &= y_n
\end{align}
or in matrix form
$$
\begin{pmatrix}
1 & x_0 & \cdots & x_0^n\\
1 & x_1 & \cdots & x_1^n\\
\vdots&&&\vdots\\
1 & x_n & \cdots & x_n^n
\end{pmatrix}
\begin{pmatrix}
a_0\\a_1\\\vdots\\a_n
\end{pmatrix}
= \begin{pmatrix}
y_0\\y_1\\\vdots\\y_n\end{pmatrix}.
$$
The $(n+1)\times(n+1)$ matrix on the left is called the Vandermonde matrix of $x_0,\ldots,x_n$, and (with some work) you can show that its determinant is always equal to
$$
\prod_{0 \leq i < j \leq n}(x_j - x_i),
$$
which is nonzero if $x_j \neq x_i$ for $j \neq i$ like we assumed for obvious reasons above. It follows that the matrix is invertible, and so there is a solution to this system of equations:
$$
\begin{pmatrix}
a_0\\a_1\\\vdots\\a_n
\end{pmatrix}
= \begin{pmatrix}
1 & x_0 & \cdots & x_0^n\\
1 & x_1 & \cdots & x_1^n\\
\vdots&&&\vdots\\
1 & x_n & \cdots & x_n^n
\end{pmatrix}^{-1}\begin{pmatrix}
y_0\\y_1\\\vdots\\y_n\end{pmatrix}.
$$
What's more, notice that we've actually proven something stronger than what we set out to. Not only is there such a polynomial of degree $n$ or less, the polynomial is unique! That is,
For any set of $n+1$ points $(x_i,y_i)$ with $x_i \neq x_j$ for $i \neq j$, there is a unique polynomial $f$ of degree at most $n$ such that $f(x_i) = y_i$ for all $i$.
If you're interested, the Wikipedia page on polynomial interpolation algorithms provides some greater technical detail about the what these coefficients look like when written out explicitly and also how efficiently some related algorithms can be implemented. This includes so-called Lagrange interpolation method, which is derived from the above solution.
Best Answer
The method you are talking about is called Lagrange Interpolation. If you want to find out the individual coefficients in the original "basis", then you would need to solve a $(n+1) \times (n+1)$ Linear system.
What we are essentially doing in Lagrange Interpolation is to shift to a new "basis" which depends on the points at which we observe the value, so that the system we need to solve is just the Identity matrix and hence we do not need to invert it as such to get the coefficients.