This is a proof by contradiction.

So suppose to the contrary,

that you can find one polynomial p as in the theorem, that has no zeros.

So no matter what you plug in there, you're not going to get 0.

Then we can look at 1 over p(z), and since p was an entire function and

it's never equal to 0, we're never dividing by 0.

Therefore, this new function f, which is 1 over p, is also an entire function.

Our goal is to apply Liouville's theorem to f.

So, we want to show that f is a bound that entire function,

thereby using Lioville's theorem making f a constant function.

And then we can conclude that p also must be constant.

So how do we do that?

We want to show that f(z) is bounded.

So we want to show f(z) is bounded above by some constant M.

We can do that by showing in turn that p(z), which is 1/f(z),

is greater than or equal to some constant bounded away from 0.

If that is the case then f(z), which is 1/p(z), is bounded above.

So that's our goal now.

Well, here's p(z) again and I rewrote it a little bit.

I just factored out a z to the n from p(z).

So here's my a n, a n-1 only had a z to the n-1, so

I have factored out too many z's.

We need to divide by z here, all the way through a0 divided by z to the n.

Now let's put absolute values on there.

The absolute value of p(z) is the absolute value of z to the n times the absolute

value of what's in parentheses.