About the linear PDE $u_{x_1x_2x_3......x_n}=u$ It is known that the linear PDE $u_{xy}=u$ has this quite nice integral kernel form of general solution e.g. according to Method of charactersitics and second order PDE.:
$u(x,y)=\int_0^xf(s)I_0\left(2\sqrt{y(x-s)}\right)~ds+\int_0^yg(s)I_0\left(2\sqrt{x(y-s)}\right)~ds$
How about the linear PDE $u_{x_1x_2x_3......x_n}=u$ ?
 A: Let $d \ge 2$ be an integer and $\vec{x}= \left( x_j \right)_{j=1}^d$.
Consider a slightly more generic PDE as given below.
\begin{equation}
\frac{\partial^d u(\vec{x})}{\prod\limits_{j=1}^d \partial x_j} = \lambda u(\vec{x}) \tag{1}
\end{equation}
Then the general solution to $(1)$ reads:
\begin{equation}
u(\vec{x}) = \sum\limits_{i=1}^d \int\limits_0^{x_i} f_i(s) \cdot {\mathfrak I}^{(d)}\left( (x_i-s) \cdot \prod\limits_{j=1,j\neq i}^d x_j\right) ds
\end{equation}
where ${\mathfrak J}^{(d)}(y):= F_{0,d-1}[-,\left(1\right)_{j=1}^{d-1};\lambda y]$  and $F_{p,q}[\cdot,\cdot;]$ is the generalized hypergeometric function.
As always, the Mathematica code snippet verifies the solution.
In[4375]:= Clear[u]; d = 5;
II[x__] := III[Times @@ x];
u[x__] := 
  Sum[Integrate[
    f[i, s] II[(x + 
        Table[If[xi == i, -s, 0], {xi, 1, Length[x]}])], {s, 0, 
     x[[i]]}], {i, 1, Length[x]}];
vecx = Table[x[i], {i, 1, d}];
u[vecx];
res = D[u[vecx], 
    Evaluate[Sequence @@ Table[{vecx[[i]], 1}, {i, 1, d}]]] // 
   Simplify;
eqs = Table[
   res[[i]] /. Integrate[Times[f[n_, s], K_], _] :> K /. 
    x[i] :> y/(Times @@ Drop[Table[x[j], {j, 1, d}], {i}]) + s, {i, 1,
     d}];
MatrixForm[eqs]
eqs1 = Tally[eqs];
DSolve[eqs1[[1, 1]] == lmb III[y], III[y], y]


