Derive the pricing formula $$P(t)=P(t,T_{n})+\sum_{i=1}^{n}[P(t,T_{i-1})-P(t,T_{i})]$$directly, by constructing a self-financing portfolio which replicates the cash flow of the floating rate bond.
$P(t,T_{i-1}) $ means Buy, at time $t$, one $T_{i−1}$-bond. This will cost $P(t, T_{i−1})$
$P(t)$ means price at $t$ time
This question is related to the Arbitrage Theory in Continuous Time book by Tomas Bjork.
In another solution, the answer is based on replication approach. Here, we provide some other approaches for the valuation of the LIBOR rate, \begin{align} L(T_{i-1}; T_{i-1}, T_i) = \frac{1}{\Delta T_i}\left(\frac{1}{P(T_{i-1}, T_i)}-1\right), \end{align} set a $T_{i-1}$ and paid at $T_i$, where $\Delta T_i =T_i-T_{i-1}$.
Let $E$ be the expectation operator under the risk-neutral measure $P$, which has the money market account value process $B_t$ as the numeraire. Then the value at time $t$ of the float payment $L(T_{i-1}; T_{i-1}, T_i)\Delta T_i$ made at $T_i$ is given by \begin{align*} B_t E\left(\frac{L(T_{i-1}; T_{i-1}, T_i)\Delta T_i}{B_{T_i}}\mid\mathcal{F}_t \right) &= B_t E\left(\frac{L(T_{i-1}; T_{i-1}, T_i)\Delta T_i}{B_{T_{i-1}}} E\left(\frac{B_{T_{i-1}}}{B_{T_i}} \mid \mathcal{F}_{T_{i-1}}\right)\mid\mathcal{F}_t \right)\\ &=B_t E\left(\frac{L(T_{i-1}; T_{i-1}, T_i)\Delta T_i}{B_{T_{i-1}}} P(T_{i-1}, T_i)\mid\mathcal{F}_t \right)\\ &=B_t E\left(\frac{1}{B_{T_{i-1}}} \Big[1 - P(T_{i-1}, T_i)\Big]\mid\mathcal{F}_t \right)\\ &= B_t E\left(\frac{1}{B_{T_{i-1}}}\mid\mathcal{F}_t \right) - B_t E\left(\frac{P(T_{i-1}, T_i)}{B_{T_{i-1}}}\mid\mathcal{F}_t \right)\\ &=P(t, T_{i-1}) - B_t \times \frac{P(t, T_i)}{B_t}\\ &= P(t, T_{i-1}) - P(t, T_i). \end{align*}
$$ $$
Alternatively, let $E_{T_i}$ be the expectation operator under the $T_i$-forward measure $P_{T_i}$, which has the bond price process $\{P(t, T_i)\mid t \geq 0\}$ as the numeraire. Then the LIBOR rate process $\{L(t; T_{i-1}, T_i) \mid 0\leq t \leq T_{i-1} \}$ is a martingale under $P_{T_i}$. Moreover, for $0 \leq t \leq T_{i-1}$, let \begin{align} \eta_t &\triangleq \frac{dP}{dP_{T_{i}}}\big|_t\\ &=\frac{B_t P(0, T_{i})}{P(t, T_i)}. \end{align} By Bayes formula, for $0 \leq t \leq T_{i-1}$, the value at time $t$ of the float payment $L(T_{i-1}; T_{i-1}, T_i)\Delta T_i$ made at $T_i$ is given by \begin{align*} B_t E\left(\frac{L(T_{i-1}; T_{i-1}, T_i)\Delta T_i}{B_{T_i}}\mid\mathcal{F}_t \right) &= B_t E_{T_i}\left(\frac{\eta_{T_i}}{\eta_t}\frac{L(T_{i-1}; T_{i-1}, T_i)\Delta T_i}{B_{T_i}}\mid\mathcal{F}_t \right)\\ &=P(t, T_i)E_{T_i}\left(L(T_{i-1}; T_{i-1}, T_i)\Delta T_i\mid\mathcal{F}_t \right)\\ &=P(t, T_i)L(t; T_{i-1}, T_i) \Delta T_i\\ &=P(t, T_{i-1}) - P(t, T_i), \end{align*} from the martingale property of $L$ under the $T_i$-forward measure $P_{T_i}$.
Edit for Gordon. First, fix point in time $T_0,...,T_n$ whereas $T_1,...,T_n$ are the coupon dates and $T_0$ is interpreted as the emission date of the bond. At time $T_i$, $i = 1,...,n$ the owner of the bond receives $c_i$.At time $T_n$ the owner receives the face value K.We now go on to compute the price of this bond, and it is obvious that the coupon bond can be replicated by holding a portfolio of zero coupon bonds with maturities $T_i$, $i = 1,...,n$.So the price,$P(t)$, at a time $t < T_1$, of the coupon bond is given by $$P(t)=KP(t,T_{n})+\sum_{i=1}^{n}c_i P(t,T_{i})$$
If the coupon rate $r_i$ is set to the spot LIBOR rate $L(T_{i−1}, T_i)$ ,then \begin{align} c_i=(T_i-T_{i−1})L(T_{i−1}, T_i)K \end{align} We now go on to compute the value of this bond at some time $t < T_0$, in the case when the coupon dates are equally spaced, with $T_i−T_{i−1}=\delta$, and to this end we study the individual coupon $c_i$. Without loss of generality we may assume that $K = 1$, and inserting the definition of the LIBOR rate \begin{align} c_i=\frac{1}{P(T_{i−1}, T_i)}-1 \end{align} The value at $t$, of the term $−1$ , is of course equal to $-P(t, T_i)$ and it remains to compute the value of the term$\frac{1}{P(T_{i−1}, T_i)}$ which is paid out at $T_i$.This is, however, easily done through the following argument.
Thus the value at $t$, of obtaining $\frac{1}{P(T_{i−1}, T_i)}$ at $T_i$, is given by $P(t, T_{i-1})$, and the value at t of the coupon $c_i$ is $P(t, T_{i−1}) − P(t, T_i)$. then, by summing up all values of $c_i$ and the value of the notional amount (i.e. 1) at $T_n$, we have $$P(t)=P(t,T_{n})+\sum_{i=1}^{n}P(t,T_{i-1})-P(t,T_{i})=p(t,T_0)$$
