Proving a number has its n-root using intermediate value theorem (most general case)

Question

I want to prove using the intermediate value theorem the existence of the n-root of a number $a$ and I'm thinking to do this using several steps:

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$n$ even

If $a>1$

Since the function $f(x)=x^n-a$ is continuous and $f(0)=-a<0$ and $f(a)=a^n-a=a(a^{n-1}-1)>0$, then by the intermediate value theorem we have the statement proved.

If $0<a<1$

Since the function $f(x)=x^n-a$ is continuous and $f(0)=-a<0$ and $f(\frac{1}{a})=\frac{1}{a^n}-a=a(\frac{1}{a^{n+1}}-1)>0$, then by the intermediate value theorem we have the statement proved.

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$n$ odd

If either $a>1$ or $0<a<1$

The same when $n$ is even

If $a<0$

I don't know

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Questions

1. Am I right so far?

2. Is there an easy way without so many steps to solve this question using Intermediate value theorem?

3. How can we prove the last part?

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