Calculating a (Forward Measure) Martingale 
Above is my question. I am, unfortunately, stuck on part (a)! Below are my workings. I've just spotted a typo -- at one point, an "$\exp$" is missing, but it's fairly obviously supposed to be there.

The only thing I can really think of is to apply Bayes' formula, but this really doesn't give me something that nice. I mean, there is a little bit of cancellation: the it with the typo cancels, but nothing else.
Any advice would be most appreciated!
As always, I'd only like assistance with the first part please! If you've seen my questions before, then you'll have seen some remark to this effect about wanting to learn maths not just be told answers, etc -- you know the drill.

Sorry for not LaTeXing it -- didn't really fancy it as it would have likely taken ages! =P
 A: I take it that you mean by "Bayes formula" the following statement:

Let $\mathbb{Q}$ be a probability measure and $\beta$ a density function. If we define a probability measure $\mathbb{P}$ by $d\mathbb{P} := \beta \, d\mathbb{Q}$, then $$\mathbb{E}_{\mathbb{P}}(X \mid \mathcal{F}) = \frac{\mathbb{E}_{\mathbb{Q}}(\beta \cdot X \mid \mathcal{F})}{\mathbb{E}_{\mathbb{Q}}(\beta \mid \mathcal{F})}$$ for any filtration $\mathcal{F}$ and $X \in L^1(\mathbb{P})$.

Here, 
$$X = \frac{P(t,\tau)}{P(t,T)} \qquad \text{and} \qquad \beta = \frac{1}{P(0,T)} e^{-\int_0^T r_u \, du}.$$
Fix $s \leq t$. Since
$$\mathbb{E}(\beta \mid \mathcal{F}_s) = \frac{1}{P(0,T)} e^{- \int_0^s r_u \, du} P(s,T)$$
we have, by Bayes' formula,
$$\begin{align*}\mathbb{E}_{\mathbb{Q}_T} \left( \frac{P(t,\tau)}{P(t,T)} \mid \mathcal{F}_s \right) &=\frac{1}{P(s,T)} e^{\int_0^s r_u \, du}\mathbb{E}_{\mathbb{Q}} \left( \frac{P(t,\tau)}{P(t,T)} e^{- \int_0^T r_u \, du} \mid \mathcal{F}_s \right) \\ &= \frac{1}{P(s,T)} \mathbb{E}_{\mathbb{Q}} \left( \frac{P(t,\tau)}{P(t,T)} e^{- \int_s^T r_u \, du} \mid \mathcal{F}_s \right).  \end{align*}$$
Consequently, we are done if we can show that
$$\mathbb{E}_{\mathbb{Q}} \left( \frac{P(t,\tau)}{P(t,T)} e^{- \int_s^T r_u \, du} \mid \mathcal{F}_s \right)  = P(s,\tau). \tag{1}$$
Using the tower property and the fact that $\frac{P(t,\tau)}{P(t,T)}$ is $\mathcal{F}_t$-measurable, we find
$$\begin{align*} \mathbb{E}_{\mathbb{Q}} \left( \frac{P(t,\tau)}{P(t,T)} e^{- \int_s^T r_u \, du} \mid \mathcal{F}_s \right) &=\mathbb{E}_{\mathbb{Q}} \left(  \mathbb{E}_{\mathbb{Q}} \left[ \frac{P(t,\tau)}{P(t,T)} e^{- \int_s^T r_u \, du} \mid \mathcal{F}_t \right] \mid \mathcal{F}_s \right) \\ &= \mathbb{E}_{\mathbb{Q}} \left( \frac{P(t,\tau)}{P(t,T)} \mathbb{E}_{\mathbb{Q}} \left[ e^{-\int_s^T r_u \, du} \mid \mathcal{F}_t \right] \mid \mathcal{F}_s \right). \end{align*}$$
As
$$\mathbb{E}_{\mathbb{Q}} \left[ e^{-\int_s^T r_u \, du} \mid \mathcal{F}_t \right] = e^{-\int_s^t r_u \, du} P(t,T)$$
this shows
$$\mathbb{E}_{\mathbb{Q}} \left( \frac{P(t,\tau)}{P(t,T)} e^{- \int_s^T r_u \, du} \mid \mathcal{F}_s \right) = \mathbb{E}_{\mathbb{Q}} \left( P(t,\tau) e^{- \int_s^t r_u \, du} \mid \mathcal{F}_s \right).$$
It follows from the pull-out property of conditional expectation that
$$P(t,\tau) e^{-\int_s^t r_u \, du} = \mathbb{E}_{\mathbb{Q}}(e^{-\int_s^{\tau} r_u \, du} \mid \mathcal{F}_t).$$
Finally, yet another application of the tower property gives
$$\begin{align*}\mathbb{E}_{\mathbb{Q}} \left( \frac{P(t,\tau)}{P(t,T)} e^{- \int_s^T r_u \, du} \mid \mathcal{F}_s \right)  &= \mathbb{E}_{\mathbb{Q}} \left[ \mathbb{E}_{\mathbb{Q}}(e^{-\int_s^{\tau} r_u \, du} \mid \mathcal{F}_t) \mid \mathcal{F}_s \right] \\ &= P(s,\tau). \end{align*}$$
