Vinogradov's "Elements of Number Theory", Ch1, Ex6: what is he talking about? This is a very specific question tied to a very specific textbook.  I wish to clarify the meaning of an exercise question in said textbook.
Exercise 6 of Chapter 1 of the 2016 Dover reprint of Saul Kravetz's 1954 English translation of "Elements of Number Theory" by I.M. Vinogradov reads:

Prove that there exist an infinite number of primes by counting the number of integers, not exceeding $N$, whose canonical decomposition does not contain prime numbers different from $p_1$, $p_2$, ..., $p_k$.

This makes it seem as though both $N$ and $\{p_1, p_2, ..., p_k\}$ should be arbitrary.  However, if both are arbitrary, then the numbers whose canonical decomposition includes only $\{p_1, p_2, ..., p_k\}$ could easily exhaust the integers not exceeding $N$.
I decided to let $\{p_1, p_2, ..., p_k\}$ be arbitrary, fixed $N = max(\{p_1, p_2, ..., p_k\})^2$, and worked from there, but I wonder if this is what Vinogradov had in mind.
A version of the textbook that appears identical to mine can be found here, and the exercise appears on page 19.  (Thanks to miracle173 for finding it.)
 A: Assume that there are only finitely many primes $p_1,\ldots, p_k$
Then there are at most 
$$\left(\frac{\log(N)}{\log(p_1)}+1\right) \cdot ... \cdot \left(\frac{\log(N)}{\log(p_k)}+1\right) \le  \left(2\frac{\log(N)}{\log(p_1)}\right) \cdot ... \cdot \left(2\frac{\log(N)}{\log(p_k)}\right) \tag{1}$$ 
numbers whose canonical decomposition includes only $\{p_1,\ldots, p_k\}$ that don't exceed $N$, if $N \ge \max\{p_1,\ldots,p_k\}$.
Therefore there are at most 
$$2^kt^k\frac{\log(N)}{\log(p_1)} \cdot ... \cdot \frac{\log(N)}{\log(p_k)} \tag{2}$$ numbers whose canonical decomposition includes only $\{p_1,\ldots, p_k\}$ that don't exceed $N^t$. But
$$\lim_{t \to \infty}2^kt^k\frac{\log(N)}{\log(p_1)} \cdot ... \cdot \frac{\log(N)}{\log(p_k)}/ N^t=0$$
A: If $Q$ is a set of primes then $$\# \{ n  \le N, \forall q \in Q, q \nmid n\} = \sum_{m \in A_Q} \mu(m) \lfloor N/m \rfloor$$
where $A_Q$ is the set of squarefree integers whose prime factors are all in $Q$ and $\mu(m)$ is the Möbius function.
Note that $\lfloor N/m \rfloor = N/m+O(1)$ so that
$$\# \{ n  \le N, \forall q \in Q, q \nmid n\} = N\sum_{m \in A_Q} \frac{\mu(m)}{m}+\sum_{m \in A_Q}\mathcal{O}(1)$$ $$= N \prod_{q \in Q} (1-\frac{1}{q}) +  \mathcal{O}(2^{|Q|}) $$
and we get that $$ \lim_{N \to \infty} \# \{ n  \le N, \forall q \in Q, q \nmid n\} =\lim_{N \to \infty} N \prod_{q \in Q} (1-\frac{1}{q}) =\infty$$ so there are infinitely many primes.
A: 
This is exactly the same answer  as https://math.stackexchange.com/a/2234659/11206 posted by of @user1952009. I made considerable changes in the presentation to try to make it more comprehensible to users that are not familiar with some notation known in number theory which were used by @user1952009.

If $P=\{p_1,\ldots,p_k\}$ is a finite  set of primes and
$$M(N):= \{n: ( n  \le N) \land ( \forall p \in P: p \nmid n)\} $$
is the set of number equal or less than $N$ that are not divided by a prime number, 
then we use the inclusion-exclusion-principle to count $M(N)$ and get
$$\begin{align}
|M(N)|&=N \\
&-\left\lfloor\frac{N}{p_1}\right\rfloor-\left\lfloor\frac{N}{p_2}\right\rfloor-\cdots-\left\lfloor\frac{N}{p_n}\right\rfloor\\
&+\left\lfloor\frac{N}{p_1p_2}\right\rfloor+\left\lfloor\frac{N}{p_1p_3}\right\rfloor+\cdots+\left\lfloor\frac{N}{p_{k-1}p_k}\right\rfloor\\
&-\left\lfloor\frac{N}{p_1p_2p_3}\right\rfloor-\left\lfloor\frac{N}{p_1p_2p_4}\right\rfloor-\cdots-\left\lfloor\frac{N}{p_{k-2}p_{k-1}p_k}\right\rfloor\\
&\vdots\\
&(-1)^k\left\lfloor\frac{N}{p_1p_2\ldots p_k}\right\rfloor\\
\end{align}$$
Note that 
$$\pm\lfloor N/m \rfloor \ge \pm N/m-1$$
so that
$$\begin{align}
|M(N)|\ge& N -1 \\
&-\frac{N}{p_1}-1-\frac{N}{p_2}-1-\cdots-\frac{N}{p_n}-1\\
&+\frac{N}{p_1p_2}-1+\frac{N}{p_1p_3}-1+\cdots+\frac{N}{p_{k-1}p_k}-1\\
&-\frac{N}{p_1p_2p_3}-1-\frac{N}{p_1p_2p_4}-1-\cdots-\frac{N}{p_{k-2}p_{k-1}p_k}-1\\
&\vdots\\
&(-1)^k\frac{N}{p_1p_2\ldots p_k}-1\\
=&N\\
&-\frac{N}{p_1}-\frac{N}{p_2}-\cdots-\frac{N}{p_n}\\
&+\frac{N}{p_1p_2}+\frac{N}{p_1p_3}+\cdots+\frac{N}{p_{k-1}p_k}\\
&-\frac{N}{p_1p_2p_3}-\frac{N}{p_1p_2p_4}-\cdots-\frac{N}{p_{k-2}p_{k-1}p_k}\\
&\vdots\\
&(-1)^k\frac{N}{p_1p_2\ldots p_k}\\
&-2^k \\
=& N \prod_{i=1}^{k} (1-\frac{1}{p_i})-2^k
\end{align}$$
If $P$ contains all primes than $M(N)=\{1\}$ and therefore $|M(N)|=1$. But 
$$\lim_{N \to +\infty}  \left(N \prod_{i=1}^{k} (1-\frac{1}{p_i})-2^k\right) =+\infty $$
and therefore if $P$ is finite we have 
$$\lim_{N \to +\infty} |M(N)| = +\infty$$
which contradicts $M(N)=\{1\}$.
