First I would like to mention that I am not too well versed with the properties of the partition function of an integer so I apologise if this question is too elementary or daft. Recall we define the partition function as $$p(n) = \text{number of ways to express $n$ as a sum of positive integers, } n\in\mathbb{Z}, n\geq 1$$ so for instance for $n=3$ we have $3 = 1+2 = 1+1+1$ so $p(3) = 3$. My question is if the following is always true for $n\geq 2$: $$\sum_{i_1 + 2i_2 + \dots=n} (-1)^{k_1+k_2+\dots}\frac{1}{1^{k_1}\cdot 2^{k_2}\cdot\dots}\frac{1}{k_1!\cdot k_2!\cdot\dots} = 0$$ Here, $k_i$ is the number of times $i$ appears in the partition. As an example, for $n=4$, we have:
$1+1+1+1+1$: $(-1)^4\cdot\frac{1}{4!} = \frac{1}{24}$
$1+1+2$: $(-1)^3\cdot\frac{1}{2}\cdot\frac{1}{2!} = -\frac{1}{4}$
$1+3$: $(-1)^2\cdot\frac{1}{3} = \frac{1}{3}$
$2+2$: $(-1)^2\cdot\frac{1}{2^2}\cdot\frac{1}{2!} = \frac{1}{8}$
$4$: is just $-\frac{1}{4}$
so the alternating sum above equals zero.
Any reference or explanation would be greatly appreciated. Thanks in advance.