Show that $2 ^{n(n+1)} \mid 32 · φ(2^{2^n} − 1) $ 
Show that for every $n \in \mathbb{N}$
$$2 ^{n(n+1)}\mid 32 · \varphi(2^{2^n}− 1)$$
where $\varphi$ denotes the Euler­-Phi function.

So the right hand side can be broken down into:
$$(2^{2^{n-1}}+1)(2^{2^{n-2}}+1)(2^{2^{n-3}}+1)\cdots(2^1 + 1)(2-1)$$
Since $φ$ is multiplicative, all that's left is to figure out the value of $\varphi(2^{2^k}+1)$ for $k$ from $1$ to $n-1$. This is where I've gotten stuck.
 A: We could use your idea of breaking $2^{2^n} - 1$ up into
$$(2^{2^{n-1}}+1)(2^{2^{n-2}}+1)(2^{2^{n-3}}+1) \ldots (2^1 + 1)(2-1)$$
Next, we could determine appropriate lower limits of each $\varphi\left(2^{2^k} + 1\right)$ and then add them together.
However, another similar (but I think a bit easier) way, with your idea being implicitly used, to prove your statement of
$$2^{n(n+1)}\,  \mid \, 32 · \varphi(2^{2^n}− 1) \tag{1}\label{eq1A}$$
is to use induction. First, using the $p$-adic order function, \eqref{eq1A} is equivalent to proving
$$n(n + 1) \le 5 + \operatorname{ord}_2\left(\varphi(2^{2^n}− 1)\right) \tag{2}\label{eq2A}$$
Start with $3$ base cases. First, with $n = 1$, we have $2 \le 5 + 1 = 6$. Next, with $n = 2$, we get $6 \le 5 + 1 + 2 = 8$. Finally, with $n = 3$, we have $12 \le 5 + 1 + 2 + 4 = 12$.
Next, assume for some $k \ge 3$ that \eqref{eq2A} is true for $n = k$. To prove \eqref{eq2A} is true for $n = k + 1$, it's sufficient to show the value of the right side increases by at least as much as the left side. To do this, first note the factoring of a difference of squares gives
$$2^{2^{k+1}} - 1 = \left(2^{2^{k}} - 1\right)\left(2^{2^{k}} + 1\right) \tag{3}\label{eq3A}$$
Since $2^{2^{k}} - 1$ and $2^{2^{k}} + 1$ differ by just $2$ and are both odd, they are also relatively prime to each other. Thus, since $\varphi$ is multiplicative, as you noted, then
$$\begin{equation}\begin{aligned}
& \varphi\left(2^{2^{k+1}} - 1\right) = \varphi\left(2^{2^{k}} - 1\right)\varphi\left(2^{2^{k}} + 1\right) \implies \\
& \operatorname{ord}_2\left(\varphi\left(2^{2^{k+1}} - 1\right)\right) = \operatorname{ord}_2\left(\varphi\left(2^{2^{k}} - 1\right)\right) + \operatorname{ord}_2\left(\varphi\left(2^{2^{k}} + 1\right)\right)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This means it's sufficient to prove
$$\operatorname{ord}_2\left(\varphi\left(2^{2^{k}} + 1\right)\right) \ge (k+1)(k+2) - k(k+1) = 2(k+1) \tag{5}\label{eq5A}$$
If $2^{2^{k}} + 1$ is prime, then $\varphi\left(2^{2^{k}} + 1\right) = 2^{2^{k}}$. It's easy to prove that $2^{k} \ge 2(k+1) \implies 2^{k-1} \ge k + 1$ for $k \ge 3$ (e.g., for $k = 3$, we get an equality, and for all larger $k$ the left side doubles but the right side increases by just $1$).
If $2^{2^{k}} + 1$ is not prime, then it must have $2$ or more prime factors. Similar to Mindlack's question comment, the Other theorems about Fermat numbers section of Wikipedia's article states and outlines a proof of a theorem by Édouard Lucas, i.e.,

Any prime divisor $p$ of $F_{n}=2^{2^{n}}+1$ is of the form $k2^{n+2}+1$ whenever $n \gt 1$

Note the How to prove that if a prime divides a Fermat Number then $p=k\cdot 2^{n+2}+1$? post also gives a proof of this result. Using this, then for each such prime $p$ factor of $2^{2^{k}} + 1$, we have
$$\operatorname{ord}_2(\varphi(p)) \ge k + 2 \tag{6}\label{eq6A}$$
Thus, for at least $2$ such prime factors, we get
$$\operatorname{ord}_2\left(\varphi\left(2^{2^{k}} + 1\right)\right) \ge 2(k + 2) \gt 2(k + 1) \tag{7}\label{eq7A}$$
This shows that, in either case, \eqref{eq5A} is true, which means that \eqref{eq2A} is true for $n = k + 1$. This proves the inductive step and, therefore, that \eqref{eq1A} is true for all $n \in \mathbb{N}$.
