Yes. Consider two cases.
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.
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. Modulo 2 we have $$\frac{m^{n-1-m}-1}{m-1}=1+m+\ldots+m^{n-m-2}\equiv 1+m(n-m-2)\equiv 1,$$ so $$ A\equiv \sum_{m=2}^{n-2} {n-2\choose m-2}=2^{n-2}-1-(n-2) $$ is indeed even.