Ramanujan's Tau function, an arithmetic property The problem:

Let $\tau(n)$ denote the Ramanujan $\tau$-function and $\sigma(n)$ be the sum of the positive divisors of $n$. Show that
  $$ (1-n)\tau(n) = 24\sum_{j=1}^{n-1} \sigma(j)\tau(n-j).$$

I'm afraid I don't even know how to start on this one.. My main problem is the $\tau$ function is defined as the coefficients of the $q$-series of the modular form $\Delta$, but that isn't (at least to my knowledge) very helpful in handling it. The properties that we have are all of a multiplicative nature (i.e., $\tau$ is multiplicative and $\tau(p^{a+2}) = \tau(p)\tau(p^{a+1}) - p^{11}\tau(p^a)$) which don't seem particularly well suited in a problem involving a sum.
Any hints, even just how to get started, would be greatly appreciated. Thanks!

Edit

As per Matt E's message, one can define, for any modular form $f$, an operator $\delta$ so that $\delta f = 12\theta f - 12E_2f$, where $\theta = q\frac{d}{dq}$. We wish to look at $\delta\Delta$. We have
\begin{align*}
 \delta\Delta &= 12\theta\Delta - 12E_2\Delta\newline
  &= 12\theta\left(\sum_{n=1}^\infty \tau(n)q^n\right) - 12E_2\Delta\newline
  &= 12q\left(\sum_{n=1}^\infty n\tau(n)q^{n-1}\right) - 12E_2\Delta\newline
  &= 12\left(\sum_{n=1}^\infty n\tau(n)q^n\right) - 12E_2\Delta.
\end{align*}
Now $\delta\Delta$ is a cusp form of weight 14, of which there are none that are non-zero. Thus we must have
$$ 12\left(\sum_{n=1}^\infty n\tau(n)q^n\right) = 12E_2\Delta.$$
Thus we may simplify a little and plug in the a series representation of $E_2$ to find
$$ \sum_{n=1}^\infty n\tau(n)q^n = (1 - 24\sum_{n=1}^\infty \sigma(n)q^n)\sum_{n=1}^\infty \tau(n)q^n.$$
Expanding the right hand side, and simplifying some more gives
$$ \sum_{n=1}^\infty n\tau(n)q^n = \sum_{n=1}^\infty \tau(n)q^n - 24\sum_{n=1}^\infty \sigma(n)q^n\sum_{n=1}^\infty \tau(n)q^n,$$
so we have
$$ \sum_{n=1}^\infty n\tau(n)q^n - \sum_{n=1}^\infty \tau(n)q^n = - 24\sum_{n=1}^\infty \sigma(n)q^n\sum_{n=1}^\infty \tau(n)q^n.$$
The end of the tunnel is starting to appear, as the left hand side looks quite nice at this point. We settle both sides into a single sum, so that we may match the coefficients on the left and right hand sides easily. We have
$$ \sum_{n=1}^\infty \tau(n)(n - 1)q^n = - 24\sum_{n=1}^\infty \left(\sum_{j=1}^{n-1} \sigma(j)\tau(n-j)q^n\right),$$
so multiplying both sides by $-1$ and matching up the coefficients gives
$$ \tau(n)(n - 1) = - 24 \left(\sum_{j=1}^{n-1} \sigma(j)\tau(n-j)\right)$$
for all $n\geq 1$, as desired.
 A: Apply the operator $\Delta$ discussed in this question to the modular form $\Delta$ (sorry for the conflict of notation vis-a-vis $\Delta$!).  The result is a modular form of weight $14$.  What can you say about it?
A: This result can be generalized :
Let $f$ be the following product :
$$f(x)=\prod_{n=1}^{+\infty}{\left(1-x^n\right)^{a_n}}=\sum_{n=0}^{+\infty}{p(n)x^n}$$
By taking the logrtihmic derivitive of $f$ we obtain :
$$\frac{f'(x)}{f(x)}=-\sum_{n=1}^{+\infty}{na_n\frac{x^{n-1}}{1-x^n}}=-\frac{1}{x}\sum_{n=1}^{+\infty}{na_n\frac{x^n}{1-x^n}}$$
with 
$$\sum_{n=1}^{+\infty}{na_n\frac{x^n}{1-x^n}}=\sum_{n=1}^{+\infty}{na_n\left(\sum_{m=1}^{+\infty}{x^{nm}}\right)}=\sum_{n=1}^{+\infty}{\left(\sum_{d|n}{da_d}\right)x^n}$$
Thus :
$$f'(x)=-\frac{f(x)}{x}\left(\sum_{n=1}^{+\infty}{\left(\sum_{d|n}{da_d}\right)x^n}\right)$$
and so
$$\sum_{n=0}^{+\infty}{np(n)x^{n}}=-\left(\sum_{n=0}^{+\infty}{p(n)x^n}\right)\left(\sum_{n=1}^{+\infty}{\left(\sum_{d|n}{da_d}\right)x^n}\right)$$
A Cauchy product gives then the recursion formula
$$p(0)=1$$
$$p(n)=\frac{-1}{n}\sum_{k=1}^n{p(n-k)\left(\sum_{d|n}{da_d}\right)}$$
(sorry form my approximative english... again)
