Let \(n\) be a natural number. If \(\dfrac{(n^{2}!)}{(n!)^{a}}\) is an integer for all \(a = 0,1,2, \ldots,b\). Find the maximum value of b in terms of n for which this is satisfied for every n. For example if b= n+8, then for n=5, \(\dfrac{(n^{2}!)}{(n!)^{a}}\) would be integer for a=1,2,3,...,b+8 i.e. 13 .

**Clarification:**

Choose the value of b so that for every natural number this would be true. There can be some numbers where b=n+9 would be true(in the example) but we have to enter b=n+8 since it is valid for every natural number.

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