Here is an approach via Langange multipliers: The Lagrangian of the constrained problem is

$$L(x,\lambda) = x_1\cdots x_n + x_2\cdots x_n + x_n - \lambda(\sum_{i=1}^n x_i - n-C).$$

The solution is a critical point of this. Taking the derivatives of $L$ with respect to all the $x_i$ you get
\begin{align}
x_2\cdots x_n -\lambda & = 0\\
x_1x_3\cdots x_n + x_3\cdots x_n - \lambda & = 0\\
\vdots\qquad & \quad \vdots\\
x_1\cdots x_{n-2}x_n + \cdots + x_n - \lambda & = 0.
\end{align}
The first equation gives 
$$\lambda = x_2\cdots x_n$$
and we multiply the further $i$th equation with $x_i$ to get
\begin{align}
x_1\cdots x_n -\lambda x_1 & = 0\\
x_1\cdots x_n + x_2\cdots x_n - \lambda x_2 & = 0\\
\vdots\qquad & \quad \vdots\\
x_1\cdots x_n + x_2\cdots x_n + x_i\cdots x_n - \lambda x_i & = 0\\
\vdots\qquad & \quad \vdots\\
x_1\cdots x_n + x_2\cdots x_n +\cdots + x_{n-1}x_n + x_n - \lambda x_n & = 0.
\end{align}
Plugging the $i$th equation into the $(i+1)$st (for $i\leq n-1$) and using $\lambda = x_2\cdots x_n$ we get 
$$
\lambda(x_i - x_{i+1}) + \frac{\lambda}{x_2\cdots x_i} = 0
$$
and 
$$
\lambda(x_{n-1}-x_n) = 0.
$$
From this I get the recursion
\begin{align}
x_2 & = x_1 + 1\\
x_3 & = x_2 + \tfrac{1}{x_2}\\
x_4 & = x_3 + \tfrac{1}{x_2 x_3}\\
\vdots & \\
x_{i+1} & = x_i + \tfrac{1}{x_2\cdots x_i}\\
\vdots & \\
x_n & = x_{n-1}.
\end{align}
(This contradicts your claim that $(x_1,\dots, x_n) = (m,m+1,\dots,m)$…)

It remains to determined $x_1$ which can in principle be done by the constraint $\sum_i x_i = n+C$, but I haven't tried…