We can prove this combinatorially, using a sign-reversing involution, also known as the the DIE method (Description, Involution, Exception).
First, let us split your sum up in a way such that all numerators are $1$. To do this, we take a sum over all ordered partitions of $n$ (also known as compositions of $n$), instead of just the integer partitions of $n$. For example, when $n=4$, this looks like
$$
-\frac1{4!}+\frac1{1!\cdot 3!}+\frac1{3!\cdot 1!}+\frac1{2!\cdot 2!}-\frac1{2!\cdot 1!\cdot1!}-\frac1{1!\cdot2!\cdot1!}-\frac1{1!\cdot1!\cdot2!}+\frac1{1!\cdot1!\cdot1!\cdot1!}
$$
Next, we multiply both sides of the equation to be proved by $n!$. The result is
$$
\sum_{\alpha \text{ is a composition of }n} \binom{n}{\alpha}(-1)^{\text{len}(\alpha)}=(-1)^n\tag1
$$
where...
$\alpha$ ranges over all lists of positive integers summing to $n$. Each $\alpha=(\alpha_1,\alpha_2,\dots,\alpha_k)$ for some $k\in \{1,\dots,n\}$, where $\alpha_1+\dots+\alpha_k=n$.
$\text{len}(\alpha)=k$.
$\binom{n}{\alpha}$ is the multinomial coefficient, defined by $$\binom n{\alpha}=\frac{n!}{\alpha_1!\alpha_2!\cdots \alpha_k!}.$$
Description
Thinking combinatorially, $\binom{n}{\alpha}$ counts the number of ways to put $n$ distinct items into $k=\text{len}(\alpha)$ distinct boxes, such that box number $i$ receives exactly $\alpha_i$ items for each $i \in \{1,\dots,k\}$. We are counting all such arrangements of $n$ items into some number of boxes, where every box is nonempty, and where arrangements with an even number of boxes count positively, and those with an odd number count negatively.
Involution
I will now define a way to divvy up all but one of these arrangements into pairs, where one of the pair counts positively, and the other counts negatively. This means that these pairs cancel themselves out in the count. Here is the method. Assume that the $n$ distinct objects are just the integers from $1$ to $n$. Let $P$ be a particular assignment of $\{1,\dots,n\}$ to some number of distinct boxes (numbered $1$ to $k$), where each box is nonempty. We give describe a different assignment, $P'$, which is the one $P$ will be paired with.
Say that number $1$ is in box number $i$. If $1$ is all by itself in $P$, then $1$ will be moved to the box on the left, and the box originally containing $1$ is deleted in $P'$ (unless $i=1$, see next bullet). Otherwise, if $i$ is in a box with other numbers, then $i$ gets kicked out to a new box on the right, numbered $i+1$ Example: $$P=\boxed{2,3},\boxed{1\phantom{,}},\boxed{4,5}\quad \iff \quad P'=\boxed{2,3,1},\boxed{4,5}$$
The only exception to this rule is when $1$ is all by itself, and is in box number $1$. In this case, we instead apply the exact same process to the number $2$. If $2$ is not alone, then kick it out to a new box on the left. If $2$ is alone, and $2$ is NOT in box number $2$, then we re-house $2$ to the box on its left.
The only exception to this is when $1$ and $2$ are both alone, and in boxes numbered $1$ and $2$ (resp.). In this case, we do the same process with number $3$.
$\vdots$
It should be clear that this process either adds exactly one box, or deletes exactly one box. Therefore, $P'$ has the opposite parity of $P$, and they cancel each other in the count.
Exception
There is exactly one arrangement which does not get a partner from this process. Namely, the arrangement with $n$ boxes, such that box number $i$ contains only the number $i$ for each $i\in\{1,\dots,n\}$. This is the only arrangement which is not canceled out, and it comes with a sign of $(-1)^n$. Therefore, we conclude that the entire sum in the LHS of $(1)$ is equal to $(-1)^n$, as desired.
