Intuition and Tricks - Crafty Short Proof - Generators, Order of a Cyclic Group - Fraleigh p. 64 Theorem 6.14 This stronger result and easier proof is based on p. 58. Hence it isn't a duplicate of this.
Theorem 206 and 207. Let $G$ be a group, $k \in \mathbb{N}$ and $a \in G$ such that $|a| = n$.
Then: 
206. $\langle a^k \rangle = \langle a^{\gcd(n,k)} \rangle$. $\qquad$
207. For any positive divisor $d$ of $n$, $|a^d | = n/d$. 

(1-2.) What's the intuition for these two theorems? The file puts them together. I understand the proof of Theorem 206. Hence I don't write it out. The original file does some examples of Theorem 207 that show to use the theorem. But they don't flesh out the intuition and I can't tumble to it. 
(3.) How do you envisage and envision the crafty first line in the proof? It feels supernormal to start with a magical exponent in $(a^d)^{n/d} = e$ that implies an inequality, $|a^d| \le n/d$. 
(4). The nice people at this explain how to envisage and envision $|a^k | = \frac{n}{gcd(n, k)}$. Is there anything to add, different, or harder about envisaging and envisoning $|a^d | = n/d$? 
 A: $1$. You are trying to find the order of the element $a^d$. The only thing you know about $a$ is that it has order $n$; in particular, $a^n = 1$. So, if you can get some power of $a^d$ to simplify to $a^n$, this will be $1$. As $d$ is a divisor of $n$, the fraction $\frac{n}{d}$ is an integer so we can take $a^d$ to the $\left(\frac{n}{d}\right)^{\text{th}}$ power which gives $$(a^d)^{\frac{n}{d}} = a^{d\times\frac{n}{d}} = a^n = 1.$$ We know that $\frac{n}{d}$ is a positive ($d$ is positive, so $\frac{n}{d}$ is positive) integer such that $(a^d)^{\frac{n}{d}} = 1$. So the order of $a^d$, which is the smallest positive integer $k$ such that $(a^d)^k = 1$, is less than $\frac{n}{d}$ (i.e. $|a^d| \leq \frac{n}{d}$). We still have to show that $\frac{n}{d}$ is the smallest such positive integer.
$2$. Here's one way to think about the result:
$$1\underset{\times a^d}{\underrightarrow{\xrightarrow{\times a} a^1 \xrightarrow{\times a}\dots \xrightarrow{\times a}}}a^d\underset{\times a^d}{\underrightarrow{\xrightarrow{\times a}\dots\xrightarrow{\times a}}}a^{2d}\underset{\phantom{\dots}\\\dots}{\xrightarrow{\times a}\dots\xrightarrow{\times a}}a^{n-d}=a^{d\left(\frac{n}{d}-1\right)}\underset{\times a^d}{\underrightarrow{\xrightarrow{\times a}\dots\xrightarrow{\times a}a^{n-1}\xrightarrow{\times a}}}1.$$
[If I were better with latex I would try to make this prettier. If anyone has suggestions of how to use latex to make this diagram clearer, please let me know.]
That is, if we follow the arrows along the top, it takes $n$ of them to get from $1$ back to $1$. As each arrow corresponds to multiplication by $a$, we see that $a^n = 1$. If we follow the arrows along the bottom, how many arrows does it take to get back to $1$? As each arrow corresponds to multiplication by $a^d$, we see that $(a^d)^? = 1$.
A: This is proved (in passing) in my answer to your prior question. The key idea is this: in $\,\Bbb Z$ we know by the Euclidean Division Algorithm and Bezout that every nonzero additive subgroup is generated by its least positive element. In particular, the Bezout identity in group theoretic language says $\, k\,\Bbb Z + n\,\Bbb Z = (k,n)\,\Bbb Z.$ $\bmod n,$ this implies $\, k\,\Bbb Z/n = (k,n)\,\Bbb Z/n,\ $ i.e. $\ \langle k\rangle = \langle (k,n)\rangle\,$ in $\,\Bbb Z/n.$
Here, again, we use $\,\Bbb Z/n,\, $ the canonical additive model of the cyclic group of order $\,n.\,$ This allows us to exploit to the hilt our well-honed intuition about integer arithmetic.
Remark $ $ A nice way to gain intuition about subgroups of cyclic groups is via toys such as Spirograph (e.g. images below) and Hoot-Nanny, which yield nice animations of the generation of subgroups of $\,\Bbb Z/n,\,$ modelled as roulette curves and/or star polygons. This is an often effective way to introduce various group-theoretical ideas to bright young children.

A: Not sure what you are asking, but here is the intuition behind 207.
Look at the group generated by $a^d$. Let $k=n/d$.
The elements are
$$a^d, a^{2d}, a^{3d}, \cdots,a^{(k-1)d}, \quad a^{kd}=1$$
The last element is $1$ since $a^n=1$. Can any earlier item be $1$? 
No, if it did then the order of the group generated by $a$ is less than $n$ (it will be no more than the power in the earlier one). Hence $\left|a^d \right| = n/d$
