I'll illustrate with an example. Let's say we have a four element set $S = \lbrace 1, 2, 3, 4 \rbrace$. We know that, given any three element set, it has $2^3 = 8$ subsets.
Let's take out our $x$ from $S$ to form a three element set $S'$. I'm going to choose $x = 2$, for no particular reason, so $S' = \lbrace 1, 3, 4 \rbrace$. Under the induction hypothesis, we assume $S'$ has $8$ subsets, which are the following:
$$\lbrace \rbrace, \lbrace 1 \rbrace, \lbrace 3 \rbrace, \lbrace 4 \rbrace, \lbrace 1, 3 \rbrace, \lbrace 3, 4 \rbrace, \lbrace 1, 4\rbrace, \lbrace 1, 3, 4 \rbrace.$$
Note that the above list are all subsets of $S$; the fact that none of them contain $x = 2$ doesn't change this. In fact these are all the sets we can form without choosing $x = 2$. We can get the rest of the sets by adding in $x = 2$ into each of the sets, to get another $8$ subsets:
$$\lbrace 2 \rbrace, \lbrace 1, 2 \rbrace, \lbrace 2, 3 \rbrace, \lbrace 2, 4 \rbrace, \lbrace 1, 2, 3 \rbrace, \lbrace 2, 3, 4 \rbrace, \lbrace 1, 2, 4\rbrace, \lbrace 1, 2, 3, 4 \rbrace.$$
Try to convince yourself that the two collections of subsets form all subsets of $S'$. Every subset of $S$ that doesn't contain $2$ corresponds uniquely with a subset of $S$ that does contain $2$. We know there are $2^3$ of the former, and due to this relationship, there must also be $2^3$ of the latter. So, in total, we have $2^3 + 2^3 = 2 \cdot 2^3 = 2^4$ subsets.
You want to show by induction that $2^n < n!$ for $n\geq 4$. Induction consists of two parts: The start and the induction step.
We start with the inductions start where we just verify that the statement is true for the lowest $n$ possible, here we have $n=4: 2^4 = 16 < 4! = 24$. That is indeed correct.
Now we prove the induction step. To not confuse the notation another variable $k$ is used. We assume that the above statement is correct for some arbitrary $k\geq 4$. We want to show that - given this condition - the statement is true for $k+1$ as well. So our goal is to show that
$$
2^{k+1}< (k+1)!
$$
After some simplification we get to the point you mentioned:
$2\cdot k! < (k+1)\cdot k!$
The inequality $2<k+1$ simply comes from the property of the arbitrary chosen $k$ which is already greater or equal to $4$. So $k+1\geq 5 > 2$ and the proof is completed.
Best Answer
Yes, the statement $P(n)$ that you prove is
and has only $n$ as free variable. Everything else, in particular $R$ is quantified within $P(n)$. For induction, you show $P(0)$, i.e.,
and you show $\forall n\colon P(n)\to P(n+1)$ to finally arrive at
For the induction step $\forall n\colon P(n)\to P(n+1)$, we proceed as follows: Let $n\in\Bbb N_0$ be arbitrary. We want to show $P(n)\to P(n+1)$. So assume $P(n)$. Now we want to show $P(n+1)$. So let $R$ be an arbitrary $n+2$ times differentiable with $R(a)=\ldots=R^{(n+1)}(a)=0$. Then $R'$ is $n+1$ times differentiable with $(R')'(a)=\ldots=(R')^{(n)}(a)=0$, hence $P(n)$ applies to $R'$ ... ... hence the conclusion holds for $R$. As $R$ was arbitrary, we conclude that $P(n+1)$ holds. Thus we have shows $P(n)\to P(n+1)$. As $n$ was arbitrary, we conclude $\forall n\in\Bbb N_0\colon P(n)\to P(n+1)$.