I am reading through Alan Baker's *Transcendental Number Theory* (don't worry if you don't know the book or the subject - this question is pretty much self-contained). Lemma 1 of Chapter 3 states an upper bound for $\nu(x;k)$, defined to be the least common multiple of the integers $x+1,\ldots,x+k$. The claim is that, for some absolute constant $c$, one has $\nu(x;k)\leq\left(\frac{c(x+k)}{k}\right)^{2k}$. The proof, as quoted from the book, is as follows:



> 
> (...) We write $\nu(x;k)=\nu'\nu''$, where all prime factors of $\nu'$, $\nu''$ are $\leq k$ and $>k$, respectively. Since the exponent to which a prime $p$ divides $\nu'$ is at most $\frac{\log(x+k)}{\log p}$, we have
>  $$\log\nu'\leq\sum\log(x+k)\leq\frac{c'k\log(x+k)}{\log k}$$
>  where the summation is over all primes $p\leq k$, and $c'$, like $c$, $c''$ and $c'''$ below, denotes an absolute constant. **Now we can assume that $k>c''$ and that $x>c''k$ for some sufficiently large $c''$, for otherwise the desired conclusion would follow at once from the simpler upper bounds $(x+k)^k$ and $c^{x+k}$ for $\nu(x;k)$. Thus we see that**
> $$\nu'\leq\left(\frac{c'''(x+k)}{k}\right)^{2k}\text{.}$$
>  But clearly $\nu''$ divides $\binom{x+k}{k}$, and this does not exceed $\frac{(x+k)^k}{k!}$; the required estimate is now apparent.
> 
> 
> 


I am really confused about the part I wrote in bold. I understand why we can assume that $k>c''$, but I don't understand why we can assume $x>c''k$; in fact, I find it strange that the "easy case" is precisely when $x$ is small compared to $k$, since this is precisely when the asserted bound does not seem useless compared to the easier (and better for $x$ large compared to $k$) bound $(x+k)^k$. Moreover, even assuming those two inequalities, I can't see how the conclusion on $\nu'$ follows.


Any help in understanding this would be appreciated; if anyone sees another way to prove this result I would also like to know it.



Upper bound on least common multiple of consecutive integers
**1.** Assume first that $k\leq c''$. Then
$$\nu(x;k)\leq(x+k)^k\leq\left(\frac{c''}{k}\right)^{2k}(x+k)^{2k}=\left(\frac{c''(x+k)}{k}\right)^{2k}.$$
So the range $k\leq c''$ is fine as long as $c$ is chosen to satisfy $c\geq c''$. 


**2.** Now assume that $x\leq c''k$. Then for some absolute constants $C$ and $C'$,
$$\nu(x;k)\leq C^{x+k}\leq C^{c''k+k}\leq C'^{2k}\leq\left(\frac{C'(x+k)}{k}\right)^{2k}.$$
So the range $x\leq c''k$ is fine as long as $c$ is chosen to satisfy $c\geq C'$.


**3.** By the above, we can assume that $k>c''$ and $x>c''k$, where $c''$ is any prescribed absolute constant. I claim that if $c''$ is sufficiently large in terms of $c'$, then $k>c''$ and $x>c''k$ imply
$$\frac{c'\log(x+k)}{2\log k}\leq\log\frac{x+k}{k}.\tag{$\ast$}$$
This is sufficient, because then the indicated bound for $\log\nu'$ implies
$$\log\nu'\leq\frac{c'k\log(x+k)}{\log k}\leq 2k\log\frac{x+k}{k},$$
which then clearly implies
$$\nu'\leq\left(\frac{x+k}{k}\right)^{2k}.$$
Let us analyze $(\ast)$. It can be rewritten as
$$\log k\leq\left(1-\frac{c'}{2\log k}\right)\log(x+k).$$
We are assuming that $x>c''k$, hence it suffices to show that
$$\log k\leq\left(1-\frac{c'}{2\log k}\right)\log(c''k+k).$$
This is equivalent to
$$\frac{c'}{2}\leq\left(1-\frac{c'}{2\log k}\right)\log(c''+1).$$
We are also assuming that $k>c''$, hence it suffices to verify that
$$\frac{c'}{2}\leq\left(1-\frac{c'}{2\log c''}\right)\log(c''+1).$$
However, it is clear that this inequality holds when $c''$ is sufficiently large in terms of $c'$. Indeed, if $c'$ is fixed and $c''$ tends to infinity, then the left hand side is fixed, while the right hand side tends to infinity. Done.


