TITLE: The number of monomials of a given degree
QUESTION [8 upvotes]: I'm trying to understand why the number of the monomials of degree $d$ in $n+1$ variables is $C_{n+d,n}$. If someone could help me to remember how to solve this, I would be glad.
Thanks.

REPLY [6 votes]: Your question confusingly lets $n$ be one less than the number of variables, a trickery to make the resulting formula more pleasant. But I don't like this kind of fiddling, because it makes me forget just which shift was necessary to get a nice answer (one more or one less?). So put $m=n+1$.
The number you are looking for, the number of monomials of degree$~d$ in the power series $\prod_{i=1}^m(1+X_i+X_i^2+X_i^3+\cdots)=\prod_{i=1}^m\frac1{1-X_i}$, is equal to the coefficient of $X^d$ in the power series $(1-X)^{-m}$ (just substitute $X_i:=X$ for all$~i$), which coefficient is by the binomial formula $(-1)^d\binom{-m}d=\binom{d+m-1}d$. Now you can again replace $m$ by $n+1$ to obtain $\binom{d+n}d=C_{n+d}^n$.
This answer assumes that you know how to "negate the upper index" in a binomial coefficient. The formula, which is well worth learning by heart, is
$$
  \binom nk=(-1)^k\binom{k-n-1}k
  \qquad\text{or equivalently}\qquad
  \binom{-n}k=(-1)^k\binom{k+n-1}k.
$$
The formula is very general, the only condition is that $k\in\Bbb N$ (and this is already assumed for the definition $\binom xk=\frac{x(x-1)\ldots(x-k+1)}{k!}$), but $n$ could be anything. 
As you can see the upper index is not purely negated, but a negative upper index is surely made non-negative. In fact the symmetry is around the interval of values $n$ for which $\binom nk=0$, namely $0\leq n<k$ (those binomial coefficients in column $k$ that stick out above Pascal's triangle, but not as far as into the negative row indexes); this symmetry is $n\mapsto k-1-n$.