Proof by Induction allows you to prove a statement about an arbitrary number n by first proving it is true when n is 1 and then assuming it is true for n=k and thereby showing it is true for n=k+1.
Let t = tan(x).
Then dt/dx = sec^2(x) = 1 + tan^2(x) = 1 + t^2
i.e. 1st derivative with 2nd power of t.
Assume that the nth derivative of tan(x) is a polynomial, in tan(x), of degree (n+1).
n = k
i.e. d^k(tan(x)) / dx^k = d^k(t) / dx^k = a0.t^(k+1) + a1.t^(k) + a2.t^(k-1) + …
We have shown above that this is true for n = 1.
Now we show that it is also true for n = k+1.
n = k+1
d^(k+1)(tan(x)) / dx^(k+1) = d(d^k(tan(x)) / dx^k) /dx
= d(d^k(t)) / dx^k) /dt * dt/dx
= d(a0.t^(k+1) + a1.t^(k) + a2.t^(k-1) + …) / dt * dt/dx
= (a0(k+1).t^k + a1k.t^(k-1) + a2(k-1).t^(k-2) + …) * (1 + t^2)
= (b0.t^k + b1.t^(k-1) + b2.t^(k-2) + …) * (1 + t^2)
= c0.t^(k+2) + c1.t^(k+1) + c2.t^k + …
i.e. i.e. (k+1)th derivative with (k+2)th power of t.
Hence, using Proof by Induction, we have shown that there is a polynomial p(u), of degree n+1, such that d^n(tan(x)) / dx^n = p(u), where u = tan(x)