How to show that a root of the equation $x (x+1)(x+2) ....... (x+2009) = c $ can have multiplicity at most 2?

Let denote by $$P(x)=x (x+1)(x+2) ....... (x+2009) - c .$$

We have $$P(0)=P(-1)=\cdots P(-2009)=-c$$ and by Rolle's theorem there's $$0>t_0>-1>\cdots>t_{2008}>-2009$$ s.t. $$P'(t_0)=P'(t_1)=\cdots=P'(t_{2008})=0.$$ Since the polynomial $P$ has the degree $2010$ then the degree of $P'$ is $2009$ and then the $t_i$ are the roots of $P'$.

If $c=0$ then the roots of $P$ are different from the roots of $P'$ and there is not a multiple root.

If $c\neq 0$, we apply the Rolle's theoem on $P'$ and we find the roots $s_i$ of $P''$ s.t $$t_0>s_0>t_1>\cdots>s_{2007}>t_{2008}$$ and if we have $P(\alpha)=P'(\alpha)=0$ then $\alpha=t_{i_0}$ ($\alpha$ is a root of $P'$) and then $\alpha\neq s_i\forall i=0,\ldots,2007$ and hence $P''(\alpha)\neq 0$.

For a more pedestrian approach involving explicit derivative computations, observe that if $c=0,$ then all roots have multiplicity $1.$ So we only consider $c\ne0.$ Let $f(x)=x(x+1)(x+2)\ldots(x+2009)$ and $g(x)=f(x)-c.$ There is a root with multiplicity greater than $2$ only if there is an $\alpha$ such that $g(\alpha)=g'(\alpha)=g''(\alpha)=0.$ But $$g'(x)=\frac{f(x)}{x}+\frac{f(x)}{x+1}+\frac{f(x)}{x+2}+\ldots\frac{f(x)}{x+2009}=\sum_{j=0}^{2009}\frac{f(x)}{x+j}$$ and $$g''(x)=\sum_{0\le j<k\le2009}\frac{2f(x)}{(x+j)(x+k)}=\frac{[g'(x)]^2}{f(x)}-\sum_{j=0}^{2009}\frac{f(x)}{(x+j)^2}.$$

If $g(\alpha)=g'(\alpha)=g''(\alpha)=0,$ then $$f(\alpha)=c,\qquad g'(\alpha)=\sum_{j=0}^{2009}\frac{c}{\alpha+j}=0,\qquad g''(\alpha)=-\sum_{j=0}^{2009}\frac{c}{(\alpha+j)^2}=0.$$ But this is impossible since all terms in the expression for $g''(\alpha)$ are non-zero of the same sign.

We have double roots for solutions $\alpha$ of the equation $$\sum_{j=0}^{2009}\frac{1}{\alpha+j}=0,$$ and the values of $c$ for which such a double root occurs are $f(\alpha).$ Note that if $\alpha$ is a double root, then so is $-2009-\alpha,$ and that the corresponding values of $c$ are equal. Hence those $c$ that give rise to double roots actually give rise to pairs of double roots.