Find the coefficient of the series

$\begin{aligned} a_0 + \frac{a_1}{n} + \frac{a_2}{n^2} + \cdots &= 1 + \frac{1}{n-1} \left (1 + \frac{1}{n-2} + \frac{1}{(n-2)(n-3)} + \cdots + \frac{1}{(n-2)!} \right ) \\ &= 1 + \frac{1}{n-1} \left (a_0 + \frac{a_1}{n-1} + \frac{a_2}{(n-1)^2} + \cdots \right ) \\ &= 1 + \left (\frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3} + \cdots \right ) \left (a_0 + a_1 \left (\frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3} + \cdots \right ) + a_2 \left (\frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3} + \cdots \right )^2 + \cdots \right ) \\ &= 1 + \frac{a_0}{n} + \frac{a_0+a_1}{n^2} + \frac{a_0+2a_1+a_2}{n^3} + \cdots \end{aligned}$ Hence we obtain $a_0 = 1$ and the recurrence $a_{m+1} = \displaystyle \sum_{k=0}^m {{m}\choose{k}} a_k$ , which matches the recurrence for the Bell numbers $B_m$ .

So $\begin{aligned} a_6 &= a_0 + 5 a_1 + 10 a_2 + 10 a_3 + 5 a_4 + a_5 \\ &= a_0 + 5 a_1 + 10 a_2 + 10 a_3 + 5 a_4 + (a_0 + 4 a_1 + 6 a_2 + 4 a_3 + a_4) \\ &= 2 a_0 + 9 a_1 + 16 a_2 + 14 a_3 + 6 (a_0 + 3 a_1 + 3 a_2 + a_3) \\ &= 8 a_0 + 27 a_1 + 34 a_2 + 20 (a_0 + 2 a_1 + a_2) \\ &= 28 a_0 + 67 a_1 + 54 (a_0 + a_1) \\ &= 82 a_0 + 121 a_1 \\ &= 203 \end{aligned}$