One way to understand how the test works is by looking at the Taylor Series of the function $f(x)$ centered around the critical point, $x = c$:
$$
f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2}(x-c)^2 + \cdots
$$
Note: In your question you said that the n-th derivative is non-zero. Here I'm assuming the n+1-st derivative is the first to be non-zero at $x=c$. It doesn't make a difference, it's just the way I learned it.
If $f'(c) = \cdots = f^{(n)}(c) = 0$ and $f^{(n+1)} \ne 0$, then the Taylor Series ends up looking like this:
$$
f(x) = f(c) + \frac{f^{(n+1)}(c)}{(n+1)!}(x-c)^{n+1} + \frac{f^{(n+2)}(c)}{(n+2)!}(x-c)^{n+2} + \cdots
$$
Consider what happens when you move $f(c)$ to the other side of the equation:
$$
f(x) - f(c) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-c)^{n+1} + \frac{f^{(n+2)}(c)}{(n+2)!}(x-c)^{n+2} + \cdots
$$
What does $f(x) - f(c)$ mean?
- If $f(x) - f(c) = 0$, then $f(x)$ has the same value as it does at $x = c$.
- If $f(x) - f(c) < 0$, then $f(x)$ has a value less than it has at $x = c$.
- If $f(x) - f(c) > 0$, then $f(x)$ has a value greater than it has at $x = c$.
We expect $f(x) - f(c) = 0$ at $x = c$ (the equation reflects this), but we're more interested in what it does on either side of $x = c$. When $x$ is really close to $c$, i.e. $(x-c)$ is a really small number, we can say:
$$
f(x) - f(c) \approx \frac{f^{(n+1)}(c)}{(n+1)!}(x-c)^{n+1}
$$
because the higher powers of a small number "don't matter" as much.
Concerning local extrema
If $n$ is odd, then our approximation of $f(x) - f(c)$ is an even-power polynomial. That means $f(x)$ has the same behavior - is either less than or greater than $f(c)$ - on both sides of $x = c$. Therefore it's a local extreme. If $f^{(n+1)}(c) > 0$, then $f(x)$ is greater than $f(c)$ on both sides of $x = c$. Otherwise, if $f^{(n+1)}(c) < 0$, then $f(x)$ is less than $f(c)$ on both sides of $x = c$
If, on the other hand, $n$ is even, then our approximation of $f(x) - f(c)$ is an odd-power polynomial centered around $x = c$. Therefore $f(x)$ will be greater than $f(c)$ on one side of $x = c$, and less on the other. That means $x = c$ isn't a local extreme.
Concerning saddle points
Note that if you differentiate both sides of our approximation twice, you get:
$$
f''(x) \approx \frac{f^{(n+1)}(c)}{(n-1)!}(x-c)^{n-1}
$$
If $n$ is even, this is another odd-power polynomial centered around $x = c$. It therefore has opposite behavior on each side of $x = c$, giving you a saddle point.
Best Answer
Let $g(x)=f^{(n-1)}(x)$.
If $g'(x)>0$ (i.e. $f(^{(n)}(x)>0$) then obviously $g(x)$ (i.e. $f^{(n-1)}(x)$) is increasing.
The correlation between the derivatives of a function is... that they are derivatives of each other. But the values of the derivatives at a single point are independent and can be quite arbitrary. For example, the derivatives of a polynomial at $0$ are simply its coefficients (times a factorial).
You will understand the meaning of the derivative test by considering the monomial functions $f(x)=x^m$ for increasing $m$: minima and inflection points alternate.