Let $P_n = a_1^n+a_2^n+a_3^n+a_4^n+a_5^n$ , where $n = 1,2,3...$ and

$S_1 = a_1+a_2+a_3+a_4+a_5$

$S_2 = a_1a_2 + a_1a_3 + a_1a_4 + a_1a_5 + a_2a_3 \\ \quad \quad + a_2a_4 + a_2a_5 + a_3a_4 + a_3a_5 + a_4a_5$

$S_3 = a_1a_2a_3 + a_1a_2a_4 + a_1a_2a_5 + a_1a_3a_4 + a_1a_3a_5 \\ \quad \quad + a_1a_4a_5 + a_2a_3a_4 + a_2a_3a_5 + a_2a_4a_5 + a_3a_4a_5$

$S_4 = a_1a_2a_3a_4 + a_1a_2a_3a_5 + a_1a_2a_4a_5 + a_1a_3a_4a_5 + a_2a_3a_4a_5$

$S_5 = a_1a_2a_3a_4a_5$

Using Newton's Identities/Sums , the relationships between $P_n$ and $S_n$ are as follows:

$\begin {cases} P_1 = S_1 & = 1 \\ P_2 = S_1P_1 - 2S_2 & = 1 \\ P_3 = S_1P_2 - S_2P_1 + 3S_3 & = 2 \\ P_4 = S_1P_3 - S_2P_2 + S_3P_1 - 4S_4 & = 3 \\ P_5 = S_1P_4 - S_2P_3 + S_3P_2 - S_4P_1 + 5S_5 & = 5 \\ P_6 = S_1P_5 - S_2P_4 + S_3P_3 - S_4P_2 + S_5P_1 & = \dfrac {m}{n} \end {cases}$

Substituting in the known values of $P_n$ , we have:

$\begin {cases} 1 = S_1 & \Rightarrow S_1 = 1 \\ 1 = 1 - 2S_2 & \Rightarrow S_2 = 0 \\ 2 = 1 - 0 + 3S_3 & \Rightarrow S_3 = \frac {1}{3} \\ 3 = 2 - 0 + \frac {1}{3} - 4S_4 &\Rightarrow S_4 = -\frac {1}{6} \\ 5 = 3 - 0 + \frac {1}{3} + \frac {1}{6} + 5S_5 & \Rightarrow S_5 = \frac {3}{10} \\ P_6 = 5 - 0 + \frac {2}{3} + \frac {1}{6} + \frac {3}{10} & = \frac {92}{15} = \frac {m}{n} \end {cases}$

$\Rightarrow m - n = 92-15 = \boxed {77}$