Find a formula for f^(100)(x) if f(x)=x^(100)

I assume you want to find the 100th derivative of f(x) = x^100

f'(x) = 100x^99 = nPr(100, 1) * x^(100 - 1)
f''(x) = 100 * 99x^98 = nPr(100, 2) * x^(100 - 2)
f'''(x) = 100 * 99 * 98x^97 = nPr(100, 23) * x^(100 - 3)
So
f⁽ⁿ⁾(x) = nPr(100, n) * x^(100 - n), for values of n = 0, 1, 2, 3, ... , 99, 100

Thus
f⁽¹ºº⁾(x) = nPr(100, 100) * x^(100 - 100)
f⁽¹ºº⁾(x) = 100! * 1
f⁽¹ºº⁾(x) = 100!

We prove by induction that f^(n)(x) = x^(10^(2n)).

It's trivially true for n=1.

Assume true for k = n-1. Then
f^(n-1)(x) = x^(10^(2n-2))

f^(n)(x) = (x^(10^(2n-2)))^100
= x^(10^2n)
QED

f^(100)(x) = x^(10^(200))

f^100(x)=(x^100)^100=x^10,000