A formula for the $r$-coloured partition function in terms of the sum of divisors function and its inverse

Let $p_{-r}(n)$ denote the $r$-coloured partition function, and $\sigma (n)={\sum }_{d\mathrm{\mid }n}d$ denote the sum of positive divisors of $n$. The aim of this note is to prove the following $$p_{-r}(n)=\theta (n)+\sum _{k=1}^{n-1}\frac{r^{k+1}}{(k+1)!}\sum _{{\alpha }_1=k}^{n-1}\sum _{{\alpha }_2=k-1}^{{\alpha }_1-1}\cdots \sum _{{\alpha }_k=1}^{{\alpha }_{k-1}-1}\theta (n-{\alpha }_1)\theta ({\alpha }_1-{\alpha }_2)\cdots \theta ({\alpha }_{k-1}-{\alpha }_k)\theta ({\alpha }_k)$$ where $\theta(n)=n^{-1}, \sigma(n)$,anditsinverse$$\sigma(n) = n,\sum_{r=1}^n \frac{(-1)^{r-1}}{r}, \binom{n}{r}, p_{-r}(n). $$