How to determine the degree of a polynomial? If 
$$g(x) = x^4 + x^3$$
From my understanding, the degree of the above polynomial i.e. $g(x)$ is 4. However, for this polynomial,
$$f(x) = (x-1)(x-2) \cdots (x-p+1)$$
What degree does $f(x)$ have? My guess was, by plug in $p = 3$, I have $f(x) = (x-1)(x-2)$, so this is polynomial of degree $p - 1$? Any idea?
Note where $p$ is prime
Thanks,
 A: There are $(p-1)$ factors, so the leading coefficient will be $x^{p-1}$, and hence the degree is indeed $p-1$.
A: As a more general guideline, for 'formal' (univariate) polynomials and polynomials over $\mathbb{R}$:


*

*The sum of a polynomial of degree $m$ and a polynomial of degree $n$ is a polynomial of degree no more than $\max(m,n)$ (but be aware that it can be less than this if $m=n$; for instance, consider the sum of $p(x) = x^3-2x^2+x+1$ and $q(x) = -x^3+2x^2+x+1$.)

*Multiplying by a (non-zero) scalar doesn't change the degree of a polynomial; the degree of $4p(x)$ is the same as the degree of $p(x)$ for any polynomial $p$.

*The product of a polynomial of degree $m$ and a polynomial of degree $n$ is a polynomial of degree exactly $m+n$ (notice how this differs from the result for sums, since no cancellation of the top-order term can happen in a product)


The last case is the one that applies to your problem; you're taking the product of $p-1$ polynomials each of degree $1$, so the degree of the result is just the sum of $p-1$ copies of $1$ - that is, $p-1$.
A: I thought I would try to answer a question raised by your comment. A version of the following lemma is true in any field, but I will formulate it for the integers modulo $p$.
Lemma Let $P(x)$ and $Q(x)$ be monic (lead coefficient $1$) polynomials of degree $d$.  If $P(a) \equiv Q(a) \pmod{p}$ for $d$ incongruent numbers $a$, then the corresponding coefficients of the two polynomials are congruent modulo $p$. 
Now look at the two polynomials $P(x)=(x-1)(x-2)\cdots(x-p+1)$ and $Q(x)=x^{p-1}-1$.  They are both monic of degree $p-1$. Clearly $P(a) \equiv 0$ for $a=1,2,\dots,p-1$. Also, $Q(a)\equiv 0$ at the same places, by Fermat's Theorem.
So corresponding coefficients are congruent to each other.  Note in particular that if $p$ is an odd prime, then the constant term of $P$ is $(p-1)!$. The constant term of $Q$ is $-1$, and we conclude that $(p-1)! \equiv -1 \pmod{p}$, Wilson's Theorem! Take any other coefficient of $P$ except the coefficient of $x^{p-1}$. The corresponding coefficient in $Q$ is $0$, which tells us that all the coefficients of $P$ are congruent to $0$ except for the first and the last.
