Note that the n-bit string, $N$, actually represents the value $\frac{N}{2^n-1}$ and we want to produce the m-bit string, $M$, where $n<m$ and
$$\frac{N}{2^n-1}\approx\frac{M}{2^m-1}$$

We scale numerator and denominator with
$$\frac{N.(2^n+1)}{(2^n-1)(2^n+1)}\approx \frac{M}{2^m-1}$$
and this simplifies to 
$$\frac{N.(2^n+1)}{2^{2n}-1}\approx \frac{M}{2^m-1}$$

In your case, you have n and m are 5 and 8 respectively and we can "stop" here. (but the process can be repeated if m >> n)

We next make the approximation:
$$\frac{N.(2^n+1)}{2^{2n}}\approx \frac{M}{2^m}$$ which simplifies to 
$$\frac{N.(2^n+1)}{2^{2n-m}}\approx M $$

Note that $N.(2^n+1)$ is equivalent to repeating the n-bit string, to create a 2n-bit string, and the division, shifts off the $2n-m$ LBS to leave an M bit result. 

QED

Of course, the  'correct' calculation is $\lfloor(\frac{(2^M-1) N}{2^n-1}+\frac{1}{2}\rfloor$ but this approximation, in general, works *most* of the time. Of course there are times when it's inaccurate, but IIRC only by one bit and relatively infrequently.