A First Course in Mathematical Analysis, David Alexander Brannan, ch 1, problem 11 Self studying Brannan's book. For some problem, I am not seeing how they got the solution.
Theorem 3. Arithmetic Mean-Geometric Mean Inequality:
For any positive real numbers $a_{1},a_{2},\dots,a_{n}$ we have
$$
(a_{1}a_{2}\dots a_{n})^{\frac{1}{n}}\leq\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}.
$$
Problem 11 Use theorem 3 with the $n+1$ positive numbers
$1,1+\frac{1}{n},1+\frac{1}{n},\dots,1+\frac{1}{n}$ to prove that,
for any positive integer $n$
$$
(1+\frac{1}{n})^{n}\leq(1+\frac{1}{n+1})^{n+1}.
$$
Book solution:
\begin{align*}
((1+\frac{1}{n})^{n})^{\frac{1}{n+1}}&\leq  \frac{1}{n+1}\times(n+1+n\times\frac{1}{n})\\
&=  \frac{n+2}{n+1}\\
&=  1+\frac{1}{n+1};
\end{align*}
taking the $(n+1)$th power of this last inequality, by the Power
Rule, we deduce that
$$
(1+\frac{1}{n})^{n}\leq(1+\frac{1}{n+1})^{n+1}.
$$
I am not seeing how Brannan got the RHS of the first line of his solution. I am having trouble seeing how it maps to theorem 3. I tried a few different things and am (embarrassingly) stuck. Can someone please explain this pedantically? Thanks.
Update: 
Here is my updated answer based on input from @Lord Shark the Unknown
(for the RHS expansion and subsequent simplification except for a
few steps at the end that I filled in). From @peek-a-boo's answer
I am gathering that (for reasons I don't understand) we are able to
expand theorem 3 to include an $(n+1)$th term, though the definition
of theorem 3 doesn't provide for one as it also doesn't provide for
the LHS having a power of $\frac{1}{n+1}$, but
rather a power of $n$. I'd like to know why we can make those changes
to theorem 3 and have it still hold (is it because of induction or ?). Anyway, here is my own updated format and solution based on their input: 
First raise both sides of the inequality to the power $\frac{1}{n+1}$
to get the inequality into a form similar to theorem 3 so that we
can use theorem 3's RHS. Then:
\begin{align*}
((1+\frac{1}{n})_{1}(1+\frac{1}{n})_{2}\cdots(1+\frac{1}{n})_{n})^{\frac{1}{n+1}}\leq & \frac{1}{n+1}\times(1+(1+\frac{1}{n})+(1+\frac{1}{n})+\cdots+(1+\frac{1}{n})) \\
= & \frac{1}{n+1}\times(1+n(1+\frac{1}{n}))\\
= & \frac{1}{n+1}\times(1+n+1)\\
= & \frac{n+2}{n+1}\\
= & \frac{n}{n+1}+\frac{2}{n+1}\\
= & 1-\frac{1}{n+1}+\frac{2}{n+1}\\
= & 1+\frac{1}{n+1}
\end{align*}
 A: The AM of $1$ and $n$ copies of $1+\frac1n$ is
$$\frac1{n+1}\left(1+\left(1+\frac1n\right)+
\cdots+\left(1+\frac1n\right)\right)=\frac1{n+1}
\left(1+n\left(1+\frac1n\right)\right)=\frac1{n+1}\left(1+n+1\right)
=\frac{n+2}{n+1}.$$
A: Let $a_1 = \dots = a_n = 1 + \dfrac{1}{n}$, and let $a_{n+1} = 1$. Now use the fact that
\begin{align}
(a_1 \dots a_{n+1})^{1/(n+1)} \leq \dfrac{a_1 + \dots + a_n + a_{n+1}}{n+1} = \dfrac{n+2}{n+1}
\end{align}
Now raise both sides to the power $n+1$.

Added after OP's edit:
In the statement of the theorem, $n$ is just an arbitrary integer $\geq 1$. So, this means it works when you have $2$ numbers $a_1,a_2$, it works if you have $7$ of them, say $a_1, a_2, \dots a_7$. So, don't try to attach any significance to the use of the symbol "$n$". It might be clearer to say the following: "given ANY FINITE COLLECTION of positive real numbers, their geometric mean is $\leq$ the arithmetic mean". So, as long as you have a finite collection of positive real numers, you can apply the theorem; it doesn't matter if it is $n$ terms or $n+10$ or $n+1$ or $n+2$ or $n-87$.
Once again, just to reiterate: suppose we are given a FINITE collection of positive real numbers. Then, the theorem says
\begin{align}
(\text{product of the given numbers})^{\frac{1}{\text{no.of numbers}}} \leq \dfrac{\text{sum of the given numbers}}{\text{no. of numbers}}
\end{align}
In this formulation, hopefully it is clear that the use of the symbol $n$ in the theorem is not the key point; it works whenever you have a finite collection of positive real numbers. Since $n$ is a finite integer, so is $n+1$, hence the theorem works if you have $n+1$ terms as well.
