First, there's a mistake or typo in the quoted statement. The requirement should be $n \geq m-1,$ not $n \geq m.$ For instance, you can fit two points $(m=2)$ with a line $(n=1).$
Second, it's most straightforward to see this with simple linear algebra, not the fundamental theorem of algebra. I'll explain below.
In general, if you have a $n$th-degree polynomial $p(x) = a_0 + a_1 x + \ldots + a_n x^n,$ and you're fitting it to the set of $m$ data points $\{ (x_1,y_1), \ldots, (x_m,y_m) \},$ then you're trying to solve the following system of linear equations:
\begin{align*}
y_1 &= p(x_1) = a_0 + a_1 x_1 + \ldots + a_n x_1^n \\
&\vdots \\
y_m &= p(x_m) = a_0 + a_1 x_m + \ldots + a_n x_m^n \\
\end{align*}
Let's write this using vectors:
\begin{align*}
\begin{bmatrix} y_1 \\ \vdots \\ y_m \end{bmatrix}
=
\begin{bmatrix} a_0 \\ \vdots \\ a_0 \end{bmatrix}
+ \begin{bmatrix} a_1x_1 \\ \vdots \\ a_1x_m \end{bmatrix}
+ \cdots
+ \begin{bmatrix} a_nx_1^n \\ \vdots \\ a_nx_m^n \end{bmatrix}
\end{align*}
And simplify a bit:
\begin{align*}
\begin{bmatrix} y_1 \\ \vdots \\ y_m \end{bmatrix}
=
a_0 \begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix}
+ a_1 \begin{bmatrix} x_1 \\ \vdots \\ x_m \end{bmatrix}
+ \cdots
+ a_n \begin{bmatrix} x_1^n \\ \vdots \\ x_m^n \end{bmatrix}
\end{align*}
In other words, a solution exists if the $y$-vector is in the span of the $x$-vectors:
\begin{align*}
\begin{bmatrix} y_1 \\ \vdots \\ y_m \end{bmatrix}
\in \textrm{span} \left \{
\begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix},
\begin{bmatrix} x_1 \\ \vdots \\ x_m \end{bmatrix},
\cdots,
\begin{bmatrix} x_1^n \\ \vdots \\ x_m^n \end{bmatrix}
\right \}
\end{align*}
Since the $y$-values are arbitrary, this is equivalent to saying that the $x$-vectors span all possible $y$-vectors in $\mathbb{R}^m{:}$
\begin{align*}
\textrm{span} \left \{
\begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix},
\begin{bmatrix} x_1 \\ \vdots \\ x_m \end{bmatrix},
\cdots,
\begin{bmatrix} x_1^n \\ \vdots \\ x_m^n \end{bmatrix}
\right \} = \mathbb{R}^m
\end{align*}
You can span $\mathbb{R}^m$ with and only with a set of $m$ independent vectors in $\mathbb{R}^m.$ So, let's count off $m$ vectors:
- The $1$st vector is $\begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix}$ which you can think of as $\begin{bmatrix} x_1^0 \\ \vdots \\ x_m^0 \end{bmatrix},$
- the components of the $2$nd vector $\begin{bmatrix} x_1 \\ \vdots \\ x_m \end{bmatrix},$ i.e. $\begin{bmatrix} x_1^1 \\ \vdots \\ x_m^1 \end{bmatrix},$ have exponents $1,$
- the components of the $3$rd vector (not shown above, but it would be $\begin{bmatrix} x_1^2 \\ \vdots \\ x_m^2 \end{bmatrix}$) have exponents $2,$
- ...
- the components of the $m$th vector have exponents $m-1.$
So, we need $n$ to be at least $m-1.$
