EDIT: I changed the answer to have it more on topic.
Summary
It boils down to Mark Joshi's answer. I wanted to add something more.
Answer
A probability measure $Q1$ and a numeraire $N1(t)$ are associated if all prices expressed relative to $N1$ are martingales under $Q1$:
$$\frac{price(t)}{N1(t)} = \mathbb{E}^{Q1} \left[ \left. \frac{price(T)}{N1(T)} \, \right| \, F_t\right] \quad \Rightarrow \quad price(t) = \mathbb{E}^{Q1} \left[ \left. \frac{N1(t) \cdot price(T)}{N1(T)} \, \right| \, F_t \right] $$
Having another probability measure $Q2$ with associated numeraire $N2(t)$, then you can change the numeraire:
$$
\begin{array}{ccl}
price(t) & = & \mathbb{E}^{Q1} \left[ \left. \frac{N1(t) \cdot price(T)}{N1(T)} \, \right| \, F_t \right] = N1(t) \cdot \mathbb{E}^{Q1} \left[ \left. \frac{price(T)}{N1(T)} \, \right| \, F_t \right] \\
& = & \mathbb{E}^{Q2} \left[ \left. \frac{N2(t) \cdot price(T)}{N2(T)} \, \right| \, F_t \right] = N2(t) \cdot \mathbb{E}^{Q2} \left[ \left. \frac{price(T)}{N2(T)} \, \right| \, F_t \right]
\end{array}
$$
Given information at time $t$, you are able to take out both numeraires at the same time since they are known.
If you set
$Q1$ equal to the risk-neutral probability measure $QRN$ and $N1(t)$ equal to the risk-free asset, i.e. $N1(t) = B(t) = e^{\int_0^t r(s) ds}$
$Q2$ equal to the T-forward probability measure $QT$ and $N2(t)$ equal to the zero-coupon bond price with maturity $T$ computed at $t$, i.e. $N2(t) = P(t,T)$
then you have the following:
$$
\begin{array}{ccl}
price(t) & = & \mathbb{E}^{QRN} \left[ \left. \frac{B(t) \cdot price(T)}{B(T)} \, \right| \, F_t \right] = B(t) \cdot \mathbb{E}^{QRN} \left[ \left. \frac{price(T)}{B(T)} \, \right| \, F_t \right] \\
& = & \mathbb{E}^{QT} \left[ \left. \frac{P(t, T) \cdot price(T)}{P(T, T)} \, \right| \, F_t \right] = P(t, T) \cdot \mathbb{E}^{QT} \left[ \left. \frac{price(T)}{P(T, T)} \, \right| \, F_t \right] \\
\end{array}
$$
The problem under the risk-neutral measure is that $B(T)$ and $price(T)$ are not independent, especially for interest rate derivatives. However, $P(T,T)$ is perfectly known: it's just $1$.
$$
price(t) = P(t, T) \cdot \mathbb{E}^{QT} \left[ \left. \frac{price(T)}{P(T, T)} \, \right| \, F_t \right] = P(t, T) \cdot \mathbb{E}^{QT} \left[ \left. price(T) \, \right| \, F_t \right]
$$
So with the change of numeraire you simplified a lot the computation, this is what you achieved.
In general, the change of numeraire is useful mainly for two reasons ($X_t$ is a process and $N_t$ is the numeraire):
$X_t = \frac{tradable asset}{N_t}$ is a martingale under the new measure, i.e. you were not able to prove directly $X_t$'s martingality but you are able under the new measure with the new numeraire.
$\frac{X_t}{N_t}$ becomes a very easy quantity to compute; this is the case of the T-forward measure in the LIBOR market model, since the expectation of the discounted payoff in the risk-neutral world becomes a deterministic discount factor times the the expectation today of the payoff.
I add that a theorem guarantees that all results are the equal: if there exists a probability measure associated to a numeraire and all assets expressed under this measure are martingales, then any other (traded) asset can be used as numeraire and any numeraire choice will lead to the same price.