[tex]n![/tex] is divisible by 4 for all [tex]n\ge4[/tex]. This means, for instance,
[tex]i^{4!} = \left(i^4\right)^{3!} = 1^{3!} = 1[/tex]
[tex]i^{5!} = \left(i^4\right)^{5\times3!} = 1^{5\times3!} = 1[/tex]
etc, so that [tex]i^{n!} = 1[/tex] for all [tex]n\ge4[/tex].
Meanwhile,
[tex]i^{0!} = i^1 = i[/tex]
[tex]i^{1!} = i^1 = i[/tex]
[tex]i^{2!} = i^2 = -1[/tex]
[tex]i^{3!} = i^6 = (-1)^3 = -1[/tex]
Then the sum we want is
[tex]i^{0!} + i^{1!} + i^{2!} + i^{3!} + 97\times1 = i + i - 1 - 1 + 97 = \boxed{95+2i}[/tex]
