Proving that there always exists an n-degree polynomial passing through any $n+1$ points I've just proven that if two $n$-degree polynomials pass through the same $n+1$ points, then they must be the same polynomial. (Otherwise, the difference of the 2 polynomials will be a non-zero ≤ n-degree polynomial with $n+1$ roots, which is not possible.
However, I don't think this implies that there can't be zero $n$ degree polynomials that pass through a given set of $n+1$ points.
Could someone please check my proof and point out any mistakes, and suggest a simpler proof if there is one?
My attempt:
Proving that there is a quadratic polynomial passing through any 3  points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$.
$$\begin{aligned}&\text{By the axioms of euclidean geometry, a straight line(linear polynomial) can be drawn through any 2 points.}\\\\ &\text{let}\ L(x) \ \text{be a linear polynomial such that}\  L(x_1)=y_1\ \text{and}\ L(x_2)=y_2\\\\ &\text {Then a quadratic polynomial}\ P(x)\ \text{can be defined as} 
\end{aligned} $$
$$P(x)=L(x)\ +\lambda(x-x_1)(x-x_2) $$ $$\text{Where λ is a suitable constant that will make}\ P(x_3)\ \text{ equal to }\ y_3$$
Continuing this argument, if there always exists a $n-1$-degree polynomial passing through any set of $n$ points, there must also be an $n$ degree polynomial passing through any set of $n+1$ points as you can define it as a sum of an $n-1$ degree polynomial passing through  $n$ of the $n+1$ points and $\lambda(x-x_1)(x-x_2)...(x-x_n) $ with suitable λ to make it pass through the $(n+1)^{th}$ point.
 A: I did not understand your question clearly, (and it might be just me and this answer may not be what you want), but are you looking for something on the means of this:
For any polynomial with the power $n$, where $n \in \mathbb N$ and requires $n + 1$ points to be solved:
A quadratic ($g(x)$) which has the $n_{th}$ power of 2 requires $n + 1$ (3) points can be defined in its general form:
$$ax^2 + bx + c$$
For every x-term there is a variable in front of it that needs to be solved and the constant (ie. a, b, c). To solve for these without any pre-given context or the use of calculus, you need 3 simulatenous equations that either relate one to another or solve for the variable itself. This means you are going to require 3 points substitute into $g(x)$ (yes, the constant included which gives the $+ 1$ in $n + 1$), where every point leads to one simultaneous equation.
For example, lets take 3 random points (4, 5), (2, 3), (6, 4):
Substituting each of the points in give:
For a:
$$5 = 16a + 4b + c$$ -->
$$a = \frac{5 - 4b - c}{16}$$
For b:
$$3 = 4a + 2b + c$$ -->
$$b = \frac{3 - 4a - c}{2}$$
For c:
$$4 = 36a + 6b + c$$ -->
$$c = 4 - 36a - 6b$$
Using these, you can create more equations by elimination or further substitution and solve for the 3 variables, to get your equation. Hope this helps!
