Many doubts about Stein & Shakarchi's Complex Analysis Question 5.2

Question

I have been working on the following problem from Stein & Shakarchi's Complex Analysis:

Find the order of growth of the following entire functions:

1. $p(z)$, $p$ is a polynomial.
2. $e^{bz^n}$, and
3. $e^{e^z}$.

For this questions I have the following references:

1. The first part of the question can be worked following the lines of @Hagen von Eitzen's answer to mse/1743986.

2. There's also this document by Robert Rhoades which references the whole problem.

And it's important to remember that Stein's definition of order of growth: $f$ has order of growth $\rho$ if

$$\rho=\inf\left\lbrace\alpha>0:\ |f(z)|\leqslant Ae^{B|z|^\alpha},\ \text{for some }A,B>0 \right\rbrace.$$

My Approach

First part

I want to be as thorough as possible with this problem so this is what I've done. Following the line of the question I've referenced, the proof for the first part goes along this lines:

First a lemma:

If $f$ and $g$ have orders of growth $\rho_f,\rho_g$ then, the order of $fg$, $\rho_{fg}$, is no larger than $\max(\rho_f,\rho_g)$

In the question the proof of this fact goes like this:

$$|fg(z)|=|f(z)||g(z)|\leq\left(A_1e^{B_1|z|^{\rho_f}}\right)\left(A_2e^{B_2|z|^{\rho_g}}\right)=A_1A_2e^{B_1|z|^{\rho_f}+B_2|z|^{\rho_g}}$$

Then in the following step I don't believe that this inequality is true:

The claim is that $$B_1|z|^{\rho_f}+B_2|z|^{\rho_g}\leq (B_1+B_2)|z|^{\max(\rho_f,\rho_g)}$$ I don't see that this true in all cases, it is true if for example we had $$B_12^{3}+B_22^{4}\leq (B_1+B_2)2^4,$$ but it isn't true for example if $$B_1\left(\frac12\right)^{3}+B_2\left(\frac12\right)^{4}\not\leq (B_1+B_2)\left(\frac12\right)^4,$$ and I don't think that the intention of this problem is to look into cases $|z|>1$, $|z|=1$ and $|z|<1$. So my first question is

Why is the inequality stated in 1743986 true?

Assuming that the lemma is true we proceed a bit differently than the answer:

Any polynomial $p$ can be factored as $$p(z)=A\prod_{k=1}^n(z-\alpha_k)$$ so by the lemma we have that $\rho_p\leq\max_{k}(\rho_k)$ where $\rho_k$ is the order of the factor $z-\alpha_k$. Analyzing a particular linear factor we have that

$$|z-\alpha|\leq|z|+|\alpha|\leq(\max(1,|\alpha|)(1+|z|).$$

Using the exponential function's Taylor series we have the inequality $1+t\leq e^t$ for all real numbers $t$. So using that and the binomial expansion we have:

$$e^t=(e^{t/n})^n\geq \left(1+\frac tn\right)^n\geq 1+\frac{t^n}{n^n},\ \text{for some }n\in\mathbb{N}.$$

Also I'm unsure about the last inequality, I know it comes from $(x+y)^n=\sum\binom{n}{k}x^ky^{n-k}$ so we take the first and last terms, but I'm unsure if it's true for all real numbers. So my second question is

Is the inequality $\left(1+\frac tn\right)^n\geq 1+\frac{t^n}{n^n}$ true for all real numbers $t$? Or is it true at least for positive $t$?

From this, taking $t=n|z|^{1/n}$ and substituting into the inequality we have

$$e^t\geq 1+\frac{t^n}{n^n}\Rightarrow e^{n|z|^{1/n}}\geq 1+\frac{(n|z|^{1/n})^n}{n^n} =1+|z|\Rightarrow e^{n|z|^{1/n}}\geq 1+|z|.$$

If I believe that the answers to the previous questions I might be able to deduce that $\rho_p\leq\frac1n$, for all $n$, but I don't believe that. This is because the $n$ in the exponent of $e^{n|z|^{1/n}}$ isn't constant!

Why can I ignore the $n$ in the exponent? If the definition treats $A,B$ as constants?

This is all I have with the first part. But for the other parts I'm confused as well

Second part

It would be natural to think that $e^{bz^n}$ has order of growth $n$ as stated in the document however I want to be very thorough with this so my approach is as follows:

$$|e^{bz^n}|=e^{\Re(bz^n)}=e^{b|z|^n\cos(n\theta)},\ \text{where }\theta=\arg(z),\ \text{the principal branch of }\arg.$$

I remember that there's an inequality that I've used for contour integrals which involves exponential functions with sines so that we may remove it. However I can't recall the inequality but my guess is that it's also valid for cosines.

Is there a way to bound $e^{b|z|^n\cos(n\theta)}$ by just $e^{b|z|^n}$?

If that was the case, then $|e^{bz^n}|\leq e^{b|z|^n}$ which means that $e^{bz^n}$ has order no greater than $n$. I am stuck trying to show that $n$ is the infimum.

If it wasn't, if we assume by contradiction that there was a value of $\rho<n$ such that $|e^{bz^n}|\leq Ae^{B|z|^\rho}\leq e^{b|z|^n}$, for some $A,B>0$, then it would be possible to work things into an inequality as follows:

$$|e^{bz^n}|\leq Ae^{B|z|^\rho}\iff e^{b|z|^n\cos(n\theta)}\leq Ae^{B|z|^\rho}\iff b|z|^n\cos(n\theta)\leq \log(A)+B|z|^\rho$$

where we have taken the real logarithm, but from here I can't quite get to anywhere after assuming $n>\rho$. I can try factoring a common $|z|^\rho$, but that leaves us with a fractional expression which is just horrible and I don't know how to manage.

How can I show succinctly that $n$ is the infimum?

Third part

For this part the claim in the document is that the order of $e^{e^z}$ is infinite. In order to show this I'd like to show that for all $n\in\mathbb{N}$ we have that

$$|e^{e^z}|\geq Ae^{B|z|^n}$$

because in that way, there would be no $Ae^{B|z|^\rho}$ to upper-bound against. So I let $n\in\mathbb N$ and start going at it as follows:

$$|e^{e^z}|=e^{\Re(e^z)}=\exp\left(e^{\Re(z)}\cos(\Im(z))\right)$$

but comparing that to $Ae^{B|z|^n}$ would take us to

$$e^{\Re(z)}\cos(\Im(z))\geq \log(A)+B|z|^n$$

which once again, I wouldn't know how to manage. So my final question is

How can I work through this part as well?

Final remarks

I know that there's equivalent definitions of order of growth, like in mse/2157304, however I prefer doing the question with Stein's original definition in the book. It should be doable using that!

Please help me be it with a comment, suggestion, partial solution or full solution. Anything will be appreciated.