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.

The estimate is best possible when $x=1$. As $\nu(1;k)=e^{\psi(k+1)}$ with $\psi$ the Chebyshev function, we have $\nu(1;k)\geq e^{c(k+1)}$.

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.

To show that this estimate is best possible, we use again $$\begin{align} \nu_p((x+1) &\cdots (x+k))-\nu_p(\nu(x;k)) \\ &=\sum_{j=1 \\p^j\leq k}^m \left(\left\lfloor \frac k{p^j}\right\rfloor+\left\{\frac k{p^j}\right\}+\left\{ \frac x{p^j}\right\}-\left\{\frac{x+k}{p^j}\right\}-1\right)\\ &\leq \sum_{j=1}^m \left\lfloor \frac k{p^j}\right\rfloor \end{align}. $$ Then $$ \nu(x;k)\geq \frac{(x+1)\cdots (x+k)}{k!}, $$ which shows that the exponent cannot be reduced below $1$.

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.

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.

The estimate is best possible when $x=1$. As $\nu(1;k)=e^{\psi(k+1)}$ with $\psi$ the Chebyshev function, we have $\nu(1;k)\geq e^{c(k+1)}$.