$N$ successive integers are linearly independent if they are large compared to $N$ Is it true for any $N\geq 1$, there is a value $f(N)$ such that for any integer $x\geq f(N)$ the integers $x+1,x+2,x+3,\ldots ,x+N$ are always multiplicatively independent (i.e. the relation $(x+1)^{e_1}(x+2)^{e_2}\ldots (x+N)^{e_N}=1$ with the $e_k\in{\mathbb Z}$ is possible only when all of the $e_k$'s are zero).
I have checked this for $N=2,3$. Let $f(N)$ be the smallest such thing described above. We have $f(2)=1$ because $x$ and $x+1$ are always coprime.
For $N=3$, we have $f(3)=2$. This can be checked as follows :
Suppose that $x$ is even. Let $2^t$ be the largest power of $2$ that divides $x+2$. Then $x+1,\frac{x+2}{2^t},x+3$ are mutually coprime odd integers and the result follows.
Next, suppose that $x\equiv 1 \ ({\mathsf{mod}}\ 4)$. Let $2^r$ be the largest power of $2$ that divides $x+3$. Then $r\geq 2$, $\frac{x+1}{2},x+2,\frac{x+3}{2^t}$ are mutually coprime odd integers and the result follows.
Finally, suppose that $x\equiv 3 \ ({\mathsf{mod}}\ 4)$. Let $2^s$ be the largest power of $2$ that divides $x+1$. Then $s\geq 2$, $\frac{x+1}{2^s},x+2,\frac{x+3}{2}$ are mutually coprime odd integers and the result follows.
 A: First, say that a number is $N$-smooth if all its prime factors are less than or equal to $N$. For example, $60=2^2\cdot 3\cdot 5$ is $5$-smooth (and also $6$-smooth, $7$-smooth, etc.)
Next, let me state a result by Tijdeman (1974): Let $S=\{p_1,p_2,...,p_k\}$ be a finite set of distinct primes, and let $a_1<a_2<a_3<...$ be the sequence of consecutive positive integers composed of primes from $S$. Then there are constants $c_1,c_2$ (depending on $k$ and $S$) such that
$$a_{n+1}-a_n\geq \frac{a_n}{c_1 (\ln a_n)^{c_2}}$$
for $n=1,2,...$ From this result, it is clear that the distance between $N$-smooth numbers goes to infinity. Simply take $S$ to be the set of primes less than or equal to $N$ and the result follows.
Now, consider the $N$ natural numbers
$$x+1,x+2,...,x+N$$
Case 1: Suppose that none of these numbers are $N$-smooth. Then there exists $p>N$ such that $p|(x+1)$. Furthermore, since $p>N$, we know $p$ does not divide any of $x+2,x+3,...x+N$. If we consider the product
$$1=(x+1)^{e_1}(x+2)^{e_2}...(x+N)^{e_N}$$
we see that $e_1=0$ as $x+1$ is the only factor $p$ divides. Of course, there is nothing special about $x+1$. The same logic applies to $x+2,x+3,...,x+N$. Thus, $e_1=e_2=...=e_N=0$.
Case 2: Suppose that some of these numbers are $N$-smooth. From the result above, we can take $x$ large enough such that the distance between consecutive $N$-smooth numbers is larger than $N$. This implies that at most one of $x+1,x+2,...,x+N$ is an $N$-smooth number. Let $x+k$ be $N$-smooth. Now, the same logic from case $1$ implies that
$$e_1=e_2=...=e_{k-1}=e_{k+1}=...=e_N=0$$
That is, every exponent is zero with the possible exception of $e_k$. This implies
$$1=(x+1)^{e_1}(x+2)^{e_2}...(x+k-1)^{e_{k-1}}(x+k)^{e_k}(x+k+1)^{e_{k+1}}...(x+N)^{e_N}$$
$$=(x+1)^{0}(x+2)^{0}...(x+k-1)^{0}(x+k)^{e_k}(x+k+1)^{0}...(x+N)^{0}=(x+k)^{e_k}$$
Since the only solution to this is $e_k=0$, we are done.
As @mathworker21 pointed out, this result by Tijdeman seems to be much stronger than is actually necessary. That is, in the original result $S$ could be any set of distinct primes while this problem simply requires $S$ to be a set of the first $k$ consecutive primes. It would be interesting to see a more elementary proof of this result. Furthermore, the linked paper proves this using a corollary to Baker's Theorem, one of the premier results in transcendental number theory. That is, it seems difficult to believe that so much mathematical machinery in transcendental number theory is necessary to solve what is a "relatively simple" problem in number theory. Of course, that's all subjective but it would be interesting to see a fully contained proof of this result that doesn't rely on such strong previous results.
A: Most certainly.

Lemma: For any $m,k \ge 1$, there is a $g(m,k)$ so that if we have $m$ positive integers $x_1 < \dots < x_m$ with $x_1 \ge g(m,k)$ and $x_m \le x_1+k-1$, then there is a prime dividing exactly one of $x_1,\dots,x_m$.

Assuming this Lemma, let's prove the existence of $f(N)$ for each $N \ge 1$. I do this by proving the existence of $f(N,k)$ for each $N,k \ge 1$, defined to be the least positive integer such that $x_1,\dots,x_N$ are multiplicatively independent if $f(N,k) \le x_1 < x_2 < \dots < x_N \le x_1+k-1$. (Note $f(N) = f(N,N)$.) We induct on $N$ to show $f(N,k)$ exists for all $k$. For $N=1$, the result is immediate. Suppose $f(N-1,k)$ exists for all $k$. Define $f(N,k) = \max(f(N-1,k),g(N,k))$. Take $x_1,\dots,x_N$ with $f(N,k) \le x_1 < x_2 < \dots < x_N \le x_1+k-1$. Then, by definition of $g(N,k)$, there is a prime $p$ dividing exactly one of $x_1,\dots,x_m$, say $x_j$. So, if $x_1^{e_1}\dots x_N^{e_N} = 1$ for $e_1,\dots,e_N \in \mathbb{Z}$, we must have $e_j = 0$. But since $y_1,\dots,y_{N-1} := x_1,\dots,x_{j-1},x_{j+1},\dots,x_N$ satisfy $f(N-1,k) \le y_1 < \dots < y_{N-1} \le y_1+k-1$, we know that $y_1,\dots,y_{N-1}$ are multiplicatively independent, completing the inductive step.

Proof of Lemma:
If $x_1 < \dots < x_m$ have $x_m \le x_1+k-1$ and no prime dividing exactly one of $x_1,\dots,x_m$, then each $x_j$ is composed only of primes that are at most $k-1$. So $x_1$ has to be small (just due to $x_1,x_m$ having prime factors of size at most $k-1$ and $x_m \le x_1+k-1$), due to a result of Tijdeman (see below).

.

Tijdeman (1974): Say a number is $k$-smooth if each of its prime factors is at most $k$. Let $a_1 < a_2 < \dots$ be all the $k$-smooth numbers. Then $\liminf_{n \to \infty} a_{n+1}-a_n = +\infty$. In fact, we have the stronger bound $a_{n+1}-a_n \ge \frac{a_n}{c_1(\log a_n)^{c_2}}$ for some positive constants depending just on $k$.

Here is a reference, found by QC_QAOA, for that Tijdeman result and its proof. I wonder if there's an easier proof for just that $\liminf_{n \to \infty} a_{n+1}-a_n = +\infty$.
