Define $a_k = (k^2 + 1)k!$ and $b_k = a_1 + a_2 + a_3 + \cdots + a_k$. Let \[\frac{a_{100}}{b_{100}} = \frac{m}{n}\] where $m$ and $n$ are relatively prime natural numbers. Find $n - m$.

Let's think step by step and output the final answer within \boxed{}.
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