Why is $\tau(n) \equiv \sigma_{11}(n) \pmod{691}$? If $n$ is a natural number, let $\displaystyle \sigma_{11}(n) = \sum_{d \mid n} d^{11}$. 
The modular form $\Delta$ is defined by $\displaystyle \Delta(q) = q \prod_{n=1}^{\infty}(1 - q^n)^{24}$.
Write $\tau(n)$ for the coefficient of $q^{n}$ in $\Delta(q)$.   
I would like to know why $\tau(n) \equiv \sigma_{11}(n) \pmod{691}$. I think the proof may be somewhat difficult, so even just an outline of the argument would be much appreciated. 
Thank you!
 A: The proof is not at all obvious if you begin simply with the formula
$$\Delta(q) = q \prod_{n=1}^{\infty} (1-q^n)^{24}.$$  However, as Derek Jennings explains in his answer, if you use the (absolutely crucial!) fact that $\Delta$ is a cusp form of weight twelve and level one, the proof is actually not very difficult.
As Derek explains, the ring of modular forms of level one is generated by two $q$-expansions, namely
$$E_4 := 1 + 240 \sum_{n = 1}^{\infty} \sigma_3(n) q^n,$$
which has weight 4,
and
$$E_6 := 1 + 504 \sum_{n=1}^{\infty} \sigma_5(n) q^n,$$
which has weight 6.  (The coefficients $240$ and $-504$ come from Bernoulli numbers, as Derek explains, but we don't need that at the moment.)
Now we see that we can make two monomials of weight 12 from these, namely
$E_4^3$ and $E_6^2$.  How do we get $\Delta$?  Well, the constant term of $\Delta$ vanishes, while $E_4^3$ and $E_6^2$ have constant term $1$, so 
$\Delta$ must be proportional to $E_4^3  - E_6^2$.  Since the coefficient
of $q$ in $\Delta$ is $1$ (i.e. $\tau(1) = 1$) while the coefficient
of $q$ in $E_4^3 - E_6^2$ is $1728,$ we find that 
$$E_4^3 - E_6^2 = 1728 \Delta.$$
It is useful to note that we can also use $E_4^3$ (say) and $\Delta$ as a basis for the weight $12$ modular forms.  In fact, they are a very convenient basis, because if $f$ is any weight $12$ modular form, with constant term $a_0$, then we can subtract of $a_0 E_4^3$ to get rid of the constant term of $f$, and then
$f - a_0 E_4^3$ must be a mulitple of $\Delta$.
To go further, we have to introduce another fact, also noted by Derek Jennings, namely that
there is a weight 12 modular form
$$E_{12} = 1 + \dfrac{65520}{691} \sum_{n=1}^{\infty} \sigma_{11}(n) q^n.$$
In fact, it is 
(for me)
easier to work with
$$691 E_{12} = 691 + 65520\sum_{n=1}^{\infty} \sigma_{11}(n) q^n,$$
which has integer coefficients.
Now we apply the above procedure to write $691 E_{12}$ in terms of $E_4^3$ and $\Delta$, to find that
$$691 E_{12} = 691 E_4^3 + (65520 - 691\cdot 720) \Delta.$$ 
Now all the $q$-expansions in this formula have integral coefficients, and so
what we find, looking at the coefficient of $q^n$, is that
$65520\sigma_{11}(n) \equiv 65520\tau(n)$ for each $n \geq 1$.
Dividing by $65520$ (which is coprime to $691$) gives the desired formula.
The usual way this is summarized is to say that
$$\dfrac{691}{65520} E_{12} = \dfrac{691}{65520} + \sum_{n=1}^{\infty} \sigma_{11}(n) q^n$$
is normalized (i.e. has coefficient of $q$ equal to $1$) and is a cuspform modulo $691$ (i.e. its constant term vanishes mod $691$).  This forces it (as we have just seen) to be congruent to $\Delta$ modulo $691$.
Finally, let me note that by far the best place to read about the theory of modular forms used here is Serre's beautiful book A course in arithmetic.
Added: Since I was editing this anyway, let me add a cultural remark, namely that the study of congruences between Eisenstein series and cuspforms,
of which the congruence between $\dfrac{691}{65520} E_{12}$ and $\Delta$ considered above is the first example, is a central topic in modern number theory.  It lies at the heart of Mazur's determination of the possible torsion subgroups of elliptic curves over $\mathbb Q$, and is the basic method via which Ribet proved his "converse to Herbrand's theorem'' result giving a criterion for non-triviality of various $p$-power subgroup of the class group of the cyclotomic field $\mathbb Q(\zeta_p)$.
