Let $f(x) = (x^n-1)/(x-1)$. Why does $f(1)=n$? In the back of De Souza (Berkeley Problems in Mathematics, page 305), it says:

For $x \neq 1$,
  $$
f(x) = (x^n-1)/(x-1) = x^{n-1} + \cdots + 1
$$
  so $f(1) = n$.

The expansion for $x \neq 1$ obviously follows from the definition of a partial geometric series. But since it requires $x \neq 1$, how can it follow that $f(1)=n$?
Edit: $f$ is defined as above as a polynomial. I think this changes the question since if $f$ is given as a polynomial, it must then be continuous. The only way for $f$ to be continuous is to define $f(1)=n$. Someone else can confirm if this is correct reasoning.

My educated guess as to the specific question you are referring to in the text is

Problem 5.1.5 (Sp77):  1.  Evaluate $P_{n-1}(1)$, where $P_{n-1}(x)$ is the polynomial $$P_{n-1}(x) = \frac{x^n-1}{x-1}.$$  2.  Consider a circle of radius $1$, and let $Q_1, Q_2, \ldots, Q_n$ be the vertices of a regular $n$-gon inscribed in the circle.  Join $Q_1$ to $Q_2, Q_3, \ldots, Q_n$ by segments of a straight line.  You obtain $(n-1)$ segments of lengths $\lambda_2, \lambda_3, \ldots, \lambda_n$.  Show that $$\prod_{i=2}^n \lambda_i = n.$$

Here I have reproduced the entirety of the problem not for the sake of answering it, but to ensure that this is the same problem that you are referencing in your question.  My source is the second edition, and the pages on which this problem appears is 59-60.  The solution (in my text) appears on page 284 and is straightforward.

All that said, I would like to comment that this problem is written rather poorly; it is oddly and imprecisely posed in part 1.  But this is itself a sort of clue as to the intention of the problem's author, for the key here is the implication that $P$, being a polynomial (and whose subscript $P_{n-1}$ is an indication of its degree), must be continuous at $x = 1$, even if its definition leads to an indeterminate form at this value.  Thus by dividing it out and equivalently expressing $P_{n-1}$ as $$P_{n-1}(x) = \sum_{i=0}^{n-1} x^i,$$ and in fact we can trivially show that this sum, multiplied by $(x-1)$, yields the numerator $x^n-1$ identically for any $x$; we therefore sidestep the indeterminate value at $x = 1$.  Again, this is just a matter of my personal opinion, but I would find it slightly objectionable to pose the problem in such a way.  I would prefer to see some language explicitly asking for a limit, for example.
