The key here, quoting from the section Infinite descent in the wikipedia article on mathematical induction, is
$\quad$ ... there are no infinite decreasing sequences of natural numbers
Here we provide constructions/hints and leave the organization/exposition of the theory to the interested reader.
Recall that we have the first projection mapping $\pi_1$ on $\Bbb Z^{+} \times \Bbb Z^{+}$ defined by:
$\quad \forall \, (m,m) \in \Bbb Z^{+} \times \Bbb Z^{+} : \pi_1(m,n)=m$
Define $P = \{ (m,n) \in \Bbb Z^{+} \times \Bbb Z^{+} \mid m \ge n \} $. Recall that the set $P$ contains the diagonal set
$\quad \quad \quad \Delta_{\mathbb Z^{+}} = \{(d,d) \mid d \in \mathbb Z^{+}\}$.
We define the function $F: P \to P$ as follows
$$
F(m,n) = \left\{\begin{array}{lr}
(m,n) & \text{if } m = n\\
(m-n,n) & \text{if } m-n \ge n\\
(n,m-n) & \text{if } m-n \lt n\\
\end{array}\right\}
$$
If $(m,n) \in P$ we can apply the $\text{gcd}$ function. Note that for elements $(d,d)$ in the diagonal $\Delta_{\mathbb Z^{+}}$,
$\tag 1 \text{gcd}(d,d) = d$
Now it is well known that
$\tag 2 \text{gcd}(m,n) = \text{gcd}\big(F(m,n)\big)$
For fixed $(s,t)$ in the domain of $F$ we define a sequence
$\tag 3 a_k = \pi_1 \circ F^k(s,t)$
By using the absurdity of an infinite descent, the sequence $(a_k)$ eventually 'stops decreasing and remains constant. That happens precisely when the algorithm $F$ 'hits the diagonal.
So the algorithm $F$ 'gets us' to the diagonal in a finite number of steps, and from there we can just 'read off' greatest common divisor.
Example: Let $m = 28$ and $n = 10$ so that $(m,n)$ belongs to the domain of $F$.
$\quad F(28,10) = (18, 10)$
$\quad F(18,10) = (10, 8)$
$\quad F(10,8) = (8, 2)$
$\quad F(8,2) = (6, 2)$
$\quad F(6,2) = (4, 2)$
$\quad F(4,2) = (2, 2)$ STOP
Of course if you don't want to stop you can continue to apply $F$. But the points on the diagonal are exactly the fixed points of $F$, so you will quickly lose interest.
The point $(2,2) \in \Delta_{\mathbb Z^{+}}$ and so $\text{gcd}(28,10) = 2$.
There’s nothing special about the fact that a factorial is involved. It’s immediate from the definition of $T$ that $T(n)=3n!$ for $n=1,2,3$; those are your base cases for this strong induction. For the induction step you simply have to use the definition of $T$ to show that if $T(k)=3k!$ for $k=1,\ldots,n-1$, where $n>3$, then $T(n)=3n!$. That definition says that
$$T(n)=T(n-3)(n^3-3n^2+2n)\,,$$
so your task is to use the induction hypothesis and a little algebra to show that the righthand side simplifies to $3n!$. HINT: It will be helpful to factor $n^3-3n^2+2n$.
Best Answer
What have you tried so far? What are you planning to use as your base case/cases and induction step?
Edit: Your idea of 1,1 is on the right track, but think of the base case as being more general. The base case can be any case where the answer is easy to test, and where you can't induct any lower. Basically these are the endpoint/edge cases,
For example, take 10 and 6. g(10,6)=g(4,6)=g(4,2)=g(2,2)=2. Another example: 15 and 5. g(15,5)=g(10,5)=g(5,5)=5.
We seem to be stopping at the g(x,y), x=y step. Which is great because that is easy to check -- of course gcd(x,x)=x.
The fact that there are three parts just means you have 3 induction steps, depending on xy, x=y. However once you get one, the other two should be easy.