LIBOR Bond Pricing

I was working through Hull's Options, Futures and other Derivatives when Hull states that the price of a Bond paying semi annual coupons in line with six month LIBOR, and discounted using a six month LIBOR discount rate is par. To me this statement seems like it's probably true but I wasn't 100%. I couldn't find the proof online, but it turns out that it is true, here is my derivation.

Proof by induction:

Base case: $n=1$
Price $= \frac{PAR + (PAR)R_c/2}{1 + R_c /2} = (PAR)\frac{1 + R_c/2}{1 + R_c /2} =$ PAR

Induction Step: Assume:

value(n year bond) $= $PAR
value(n+1 year bond) = $\sum\limits_{i=1}^{2(n+1)-1}\frac{(PAR)R/2}{(1+R/2)^i} + \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}}$

Breaking the summation down to n years and adding in zero:

$ = \sum\limits_{i=1}^{2n-1}\frac{(PAR)R/2}{(1+R/2)^i} + \sum\limits_{i=2n}^{2(n+1)-1}\frac{(PAR)R/2}{(1+R/2)^i}$ $+ \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}} + \left(\frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}-\frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}\right)$

Rearranging:

$ = \left(\sum\limits_{i=1}^{2n-1}\frac{(PAR)R/2}{(1+R/2)^i} + \frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}\right) + \sum\limits_{i=2n}^{2(n+1)-1}\frac{(PAR)R/2}{(1+R/2)^i}+ \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}} -\frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}$

Then using the fact that a n year bond is priced at PAR:

$ = PAR + \sum\limits_{i=2n}^{2(n+1)-1}\frac{(PAR)R/2}{(1+R/2)^i}+ \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}} -\frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}$

No we just take the PAR outside and cancel out to 1.

$ = PAR\left(1 + \sum\limits_{i=2n}^{2(n+1)-1}\frac{R/2}{(1+R/2)^i}+ \frac{(1+R/2)}{(1+R/2)^{(2(n+1))}} -\frac{(1+R/2)}{(1+R/2)^{2n}}\right)$

$ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \frac{R/2}{(1+R/2)^(2n+1)}+ \frac{1}{(1+R/2)^{(2n+1)}} -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \left(\frac{R/2}{(1+R/2)^(2n+1)}+ \frac{1}{(1+R/2)^{(2n+1)}}\right) -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \frac{1+ R/2}{(1+R/2)^(2n+1)}+ -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \frac{1}{(1+R/2)^(2n)}+ -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR\left(1 + \left(\frac{R/2}{(1+R/2)^{(2n)}} + \frac{1}{(1+R/2)^(2n)} \right)+ -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR\left(1 + \frac{1 + R/2}{(1+R/2)^{(2n)}} -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR\left(1 + \frac{1}{(1+R/2)^{(2n -1)}} -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR$