Yes. Consider two cases.

1) Assume that $n$ is even. Then $$c_n=1+\sum_{m=1}^{n-1}m{n\choose m}m^{n-1-m}=c_n=1+\sum_{m=1}^{n-1}n{n-1\choose m-1}m^{n-1-m}$$
is indeed odd.

2) Let $n>1$ be odd. Then $$c_n=1+n\sum_{m=1}^{n-1}{n-1\choose m-1}+n\sum_{m=2}^{n-2}{n-1\choose m-1}(m^{n-1-m}-1)\\=1+n(2^{n-1}-1)+n(n-1)\sum_{m=2}^{n-2}{n-2\choose m-2}\cdot \frac{m^{n-1-m}-1}{m-1}\\=:1+n(2^{n-1}-1)+n(n-1)A.$$
It already follows that $\nu_2(c_n)\geqslant \nu_2(n-1)$, since $c_n\equiv n\cdot 2^{n-1} \pmod {n-1}$, and $\nu_2(n-1)<n-1$. The equality $\nu_2(c_n)=\nu_2(n-1)$ is thus equivalent to $A$ being even. Denote $n-2=s$ (so $s$ is odd) and $m-2=j$, we get
$$
A=\sum_{j=0}^{s-2} {s\choose j}\cdot \frac{(j+2)^{s-j-1}-1}{j+1}.
$$
Modulo 2 we have $$\frac{(j+2)^{s-j-1}-1}{j+1}=1+(j+2)+\ldots+(j+2)^{s-j-2}\equiv 1+(j+2)(s-j-2)\equiv 1,$$
so 
$$
A\equiv \sum_{j=0}^{s-2} {s\choose j}=2^{s-2}-s-1
$$
is even.