Yes, this is true. The idea is to get $n-1$ out of the sum as a multiple.

We have $$c_n=1+\sum_{m=1}^{n-1}m{n\choose m}m^{n-1-m}=c_n=1+\sum_{m=1}^{n-1}n{n-1\choose m-1}m^{n-1-m}$$
We see that if $n$ is even, then $c_n$ is odd.

Now let $n>1$ be odd. Then proceed this way:$$c_n=1+n\sum_{m=1}^{n-1}{n-1\choose m-1}+n\sum_{m=2}^{n-2}{n-1\choose m-1}(m^{n-1-m}-1)\\=1+n(2^{n-1}-1)+n(n-1)\sum_{m=2}^{n-2}{n-2\choose m-2}\cdot \frac{m^{n-1-m}-1}{m-1}\\=:1+n(2^{n-1}-1)+n(n-1)A.$$
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 
$$
A\equiv \sum_{m=2}^{n-2} {n-2\choose m-2}=2^{n-2}-1-(n-2)\equiv 0\pmod 2.
$$
Then $c_n=(1-n)+(n2^{n-1})+n(n-1)A$, and $1-n$ has 2-adic valuation strictly smaller than two other summands, thus $\nu_2(c_n)=\nu_2(1-n)$.