Alternatively, we can prove a stronger bound as stated in Baker's book: **The exponent $2$ can in fact be reduced easily to $1$, which is best possible**. 

We compare the exponents of a prime $p$ in $(x+1)\cdots (x+k)$ and $\nu(x;k)$. Note that 
$$\left\lfloor \frac{x+k}{p^j}\right\rfloor-\left\lfloor \frac x{p^j}\right\rfloor=\left\lfloor \frac k{p^j}\right\rfloor+\left\{\frac k{p^j}\right\}+\left\{ \frac x{p^j}\right\}-\left\{\frac{x+k}{p^j}\right\}$$
counts the number of multiples of $p^j$ in $x+1, \ldots, x+k$. The last three fractional part terms are either $0$ or $1$. 

Let $ap^m$ with $(a,p)=1$ be the largest $p$-power appears in $x+1,\ldots x+k$. Then we expect there is more $p$-power in the product $(x+1)\cdots (x+k)$ than in $\nu(x;k)$. The excess $p$-power is
$$\begin{align}
\nu_p((x+1) &\cdots (x+k))-\nu_p(\nu(x;k)) \\
&=\sum_{j=1}^m \left(\left\lfloor \frac{x+k}{p^j}\right\rfloor-\left\lfloor \frac x{p^j}\right\rfloor-1 \right)\\
&=\sum_{j=1 \\p^j\leq k}^m \left(\left\lfloor \frac k{p^j}\right\rfloor+\left\{\frac k{p^j}\right\}+\left\{ \frac x{p^j}\right\}-\left\{\frac{x+k}{p^j}\right\}-1\right)\\
& \ \ \ +\sum_{j=1 \\p^j> k}^m \left(\left\lfloor \frac{x+k}{p^j}\right\rfloor-\left\lfloor \frac x{p^j}\right\rfloor-1 \right)\end{align}.
$$
If $p^j>k$, then $ap^m$ is the unique multiple of $p^j$ in $x+1,\ldots, x+k$. Thus, the difference between the floor functions in the second sum is $1$. Then, the second sum must vanish. 

Using that the three fractional parts in the first sum is either $0$ or $1$, we have 
$$\begin{align}
\nu_p((x+1) &\cdots (x+k))-\nu_p(\nu(x;k)) \\
&=\sum_{j=1 \\p^j\leq k}^m \left(\left\lfloor \frac k{p^j}\right\rfloor+\left\{\frac k{p^j}\right\}+\left\{ \frac x{p^j}\right\}-\left\{\frac{x+k}{p^j}\right\}-1\right)\\
&\geq \sum_{j=1 \\p^j\leq k}^m \left(\left\lfloor \frac k{p^j}\right\rfloor -1\right) \geq \sum_{j=1}^m \left\lfloor \frac k{p^j}\right\rfloor - \frac{\log k}{\log p}.
\end{align}
$$
Hence, 
$$
\nu(x;k)\leq \frac{(x+1)\cdots (x+k)}{\prod_{p\leq k} \left( p^{\sum_{j=1}^m \left\lfloor \frac k{p^j}\right\rfloor - \frac{\log k}{\log p}}\right)}\leq \frac{(x+1)\cdots (x+k)}{k!} \cdot \prod_{p\leq k} k.
$$
By Mertens' estimate, $\prod_{p\leq k} k\leq k^{ck/\log k}=e^{ck}$. 
Therefore, the desired estimate
$$
\nu(x;k)\leq \frac{(x+k)^ke^{ck}}{k!} \leq \left(\frac{c'(x+k)}k\right)^k
$$
follows.