Larger focal length lenses do have less optical power, as you have stated.
A single lens, however, does not form a telescope. You need two lenses, e.g. an objective lens and an eye lens with respective focal lengths $f_\text{eye}$ and $f_\text{obj}$. The magnification of a telescope constructed with these two lenses will have a magnification
$$ M = \left|\frac{f_\text{obj}}{f_\text{eye}}\right|$$
A larger objective focal length does result in larger magnification -- however, the individually higher optical power element in this telescope is actually the eye lens.
To achieve large magnification, you need either a really short focal length eye lens or a really long focal length objective. It is much more difficult to manufacture high quality short focal length lenses than it is to manufacture high quality long focal length lenses. This is because short focal lenses are more curved. In the thin lens approximation, the focal length of a lens of index $n$ in air with radii of curvature $R_1,R_2$ is given by:
$$ \frac{1}{f} = (n-1)\left( \frac{1}{R_1} - \frac{1}{R_2}\right)$$
As a lens focal length gets longer and longer, the lens becomes less and less curved. An infinite focal length lens would be a pane of glass, or a window.
Can you use a "window" as your objective lens to achieve infinite magnification? In principle, yes, however, this is not a practical solution primarily because the length of a two-lens telescope is $L = f_\text{obj} + f_\text{eye}$. In other words, such a telescope would need to be infinite in length.
These are all good questions
So, I'm guessing, the resultant focal length of the system (whatever that is), will also depend upon the distance between the lenses. How does the power of the lens fit into all this?
That is correct. In the limit where your lenses are thin the powers of two lenses is added as follows:
$$\phi_\text{tot} = \phi_1 + \phi_2 - \phi_1\phi_2\tau$$
where the individual powers are given by $\phi_1 = 1/f_1,\phi_2 = 1/f_2$, and $\tau = t/n$. $t$ is the spacing between the two lens elements and $n$ is the index of refraction of the medium between your two lenses ($n=1$ in vacuum). The effective focal length is then $f_\text{effective} = 1/\phi_\text{tot}$
There is a distinction between the effective focal length $f_\text{effective}$, the rear focal length $f_\text{R}$, and the front focal length $f_\text{F}$.
$$f_\text{R} = n_\text{R}f_\text{effective}\quad \quad f_\text{Front} = -n_\text{Front}f_\text{effective}$$
I.e. the front focal length is measured from the front principal plane to the front focal point (negative distance for a positive effective focal length) and the rear focal length is measured from the rear principal plane to the rear focal point (positive distance for a positive effective focal length).
Also, what do they mean by the 'resultant power of the lenses' in the first place? The resultant system of lenses is not going to be like these lenses which have negligible thickness. Where are you going to measure the focal length from?
This is a very important point. For a system of multiple lenses you measure the front and rear focal lengths from what are called the front and rear principal planes (rather than measuring directly from a lens element). For the case of a single lens the front and rear principal planes are located at the lens. For two lenses, the front principal plane is shifted from the first lens element a distance
$$d_\text{F} = + \frac{\phi_1}{\phi_\text{tot}} \frac{n_\text{F}\;t}{n}$$
the rear principal plane is shifted from the second lens element a distance.
$$d_\text{R} = -\frac{\phi_1}{\phi_\text{tot}}\frac{n_\text{R}\;t}{n}$$
For the above two equations I am defining the first lens to be on the left and the second lens on the right. The separation is $t$ and the index of the medium between the two lenses is $n$. The index of the medium to the right of the two lenses is $n_\text{R}$ and the index of the medium to the left of the two lenses is $n_\text{F}$. A value of $d_\text{F}$ or $d_\text{R}$ less than zero indicates a shift of the respective principal plane to the left; a value greater than zero indicates a shift to the right.
Disclaimer: this response is only a very brief introduction to the ideas of gaussian optics and the cardinal points. This stuff can be really confusing for anybody, especially when you consider all of the sign conventions. Also, these equations are really only valid in the paraxial limit for rotationally symmetric systems. Having said that, these basic formulations can be expanded to a system of any number of lenses -- not just 2. If you really want to understand this stuff there are surely some good books on this material. Hecht seems to be the book for intro optics, although I haven't read him. Check the table of contents to make sure gaussian optics is covered (it should be) because that's quite an expensive text.
Best Answer
It's like leverage. The longer the distance from the objective lens to the virtual image, the larger the virtual image.
Imagine there's a piece of frosted glass at the focal point. It will show the virtual image.
Now the eyepiece looks at that virtual image with a magnifying glass. That also makes it look bigger.