Differential equations of the form $M\,dx+N\,dy=0$ such that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ are said to be exact, because the left hand side of our equation is the exact differential of some function f, and the DE has a solution of the form $f(x,y)=c$, where c is a constant. Standard textbooks give us a procedure for solving such equations that essentially amounts to using the identity $f=\int\frac{\partial f}{\partial x}\: dx+\int(\frac{\partial f}{\partial y}-\frac{\partial}{\partial y}\int\frac{\partial f}{\partial x}\: dx)\: dy$. Specifically, we integrate M (respectively N, &c.) with respect to x to get $f=\int\frac{\partial f}{\partial x}\: dx+C(y)$, and then differentiate this with respect to y and do algebra to find C'(y), which we can then integrate to find C(y), which we can substitute back into our expression for f.
It occurred to me that this technique should generalize---that is, when instead of $M\,dx+N\,dy=0$ such that $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$, we have $\sum_{i}M_{i}\,dx_{i}=0$ where the $M_{i}$ satisfy the criteria for being an exact differential, then we can iteratively apply the same procedure to solve the equation. And so I came up with the following---
Theorem (?). A differential equation of the form $\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}=0$ for some function f has an implicit solution of the form $f=\sum_{i=1}^{n}a_{i}=c$ where $a_{0}=0$, $a_{i}=\int\left(\frac{\partial f}{\partial x_{i}}-\frac{\partial}{\partial x_{i}}\left(\sum_{j=1}^{i-1}a_{j}\right)\right)\: dx_{i}$ for $i\in\mathbb{N}_{+}$, and c is an arbitrary constant.
Proof (?). By induction on the number of variables n. The theorem is true for n=1 because $f=\int\,(\frac{\partial f}{\partial x}-0)\,dx$ by the fundamental theorem of calculus.
To complete the induction, we need to show that if $f(x_{1},...,x_{n})=\sum_{i=1}^{n}a_{i}=c$ is a solution to $\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}=0$, then $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n+1}a_{i}=c$ is a solution to $\sum_{i=1}^{n+1}\frac{\partial f}{\partial x_{i}}=0$. Think of $f(x_{1},...,x_{n})$ as being the "special case" of $f(x_{1},...,x_{n+1})$ where $x_{n+1}$ is being "treated like a constant." (My thinking here is not nearly as precise as it should be and I'm overloading the function name f, but hopefully you understand the intuition to which I am appealing.) So we can say that $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n}a_{i}+C(x_{n+1})$.
Then we proceed analogously as in the two-variable case. Differentiating by $x_{n+1}$ and applying algebra yields $C'(x_{n+1})=\frac{\partial f}{\partial x_{n+1}}-\frac{\partial}{\partial x_{n+1}}\left(\sum_{i=1}^{n}a_{i}\right)$, and then integrating with respect to $x_{n+1}$ yields $C(x_{n+1})=\int\left(\frac{\partial f}{\partial x_{n+1}}-\frac{\partial f}{\partial x_{n+1}}\left(\sum_{i=1}^{n}a_{i}\right)\right)\: dx_{n+1}$.
Then substituting into our earlier expression for $f(x_{1},...,x_{n+1})$ we get $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n}a_{i}+\int\left(\frac{\partial f}{\partial x_{n+1}}-\frac{\partial f}{\partial x_{n+1}}\left(\sum_{i=1}^{n}a_{i}\right)\right)\: dx_{n+1}$, which by virtue of the definition of $a_{i}$ is equivalent to $f(x_{1},...,x_{n+1})=\sum_{i=1}^{n+1}a_{i}$. But by the principle of induction, this is quod erat demonstrandum.
Example. In the n=3 case (and naming our variables x, y, and z), we get $f(x,y,z)=\int\frac{\partial f}{\partial x}\: dx+\int(\frac{\partial f}{\partial y}-\frac{\partial}{\partial y}\int\frac{\partial f}{\partial x}\: dx)\: dy$ $+\int(\frac{\partial f}{\partial z}-\frac{\partial}{\partial z}(\int\frac{\partial f}{\partial x}\: dx+\int(\frac{\partial f}{\partial y}-\frac{\partial}{\partial y}\int\frac{\partial f}{\partial x}\: dx)\: dy))\, dz$, which is seen to be an identity by performing the integrations. What's going on is that $a_{1}=\int\frac{\partial f}{\partial x}\: dx=f$, and all following terms are zero by design, e.g., $a_{2}=f-f$.
So my question is (and I offer my sincerest apologies if this is not an appropriate question, or if my poor exposition has rendered my intentions virtually unreadable) does this seem basically correct, or am I doing something wrong? And how do I fix my sloppy reasoning in the inductive step ("special case" ... "treated like a constant")? I would be much obliged for any input.