Can we parametrize the Fermat curve by $g(t)^n+g'(t)^n=1$ for already named $g(t)$? Basically the Fermat curve
is $x^n+y^n=1$.
For $n > 2$, can we parametrize it by $x=g(t),y=g'(t)$ for already named function $g(t)$ of complex argument?
i.e., $g(t)^n+g'(t)^n=1$.

Some of what I know:
For $n=2$, solution is $g(t)=\sin(t)$.
For $n=4$, Wolfram alpha appears to give functional equation.
For $n \ge 3$, SAGE gives functional equation with a lot of constants:
#n = 3
x == 1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-t^3 + 1)^(1/3)/t - 1)) + _C + 1/3*log((-t^3 + 1)^(1/3)/t + 1) - 1/6*log(-(-t^3 + 1)^(1/3)/t + (-t^3 + 1)^(2/3)/t^2 + 1),
g(x) == (-t^3 + 1)^(1/3)
#n = 5
x == 1/5*sqrt(5)*(sqrt(5) + 1)*arctan((sqrt(5) + 4*(-t^5 + 1)^(1/5)/t - 1)/sqrt(2*sqrt(5) + 10))/sqrt(2*sqrt(5) + 10) + 1/5*sqrt(5)*(sqrt(5) - 1)*arctan(-(sqrt(5) - 4*(-t^5 + 1)^(1/5)/t + 1)/sqrt(-2*sqrt(5) + 10))/sqrt(-2*sqrt(5) + 10) + _C - 1/10*(sqrt(5) + 3)*log(-(-t^5 + 1)^(1/5)*(sqrt(5) + 1)/t + 2*(-t^5 + 1)^(2/5)/t^2 + 2)/(sqrt(5) + 1) - 1/10*(sqrt(5) - 3)*log((-t^5 + 1)^(1/5)*(sqrt(5) - 1)/t + 2*(-t^5 + 1)^(2/5)/t^2 + 2)/(sqrt(5) - 1) + 1/5*log((-t^5 + 1)^(1/5)/t + 1),
g(x) == (-t^5 + 1)^(1/5)

 A: Let me give a perspective extending what Geoff did. He notes the function is the inverse function of the integral 
$$\int_0^t \frac{1}{(1-u^n)^{1/n}} du $$
with the change of variables $v=u^n$ so $dv= n u^{n-1} du = n v^{(n-1)/n} dv$
$$= \frac{1}{n} \int_0^{t^{1/n}} \frac{1}{ (1-v)^{1/n} v^{(n-1)/n}} dv$$
we recognise the important special case $t=1$ of this integral as $1/n$ times the beta function $\beta((n-1)/n, 1/n)$.
In general the integral from $0$ to $t$ may depend on the choice of contour. The dependence is governed by a sheaf cohomology group, which here is one-dimensional. This should imply that the integral is well-defined up to the period $\beta((n-1)/n,1/n)$ times an element of $\mathbb Z[\mu_n]$.
In particular for $n \leq 4$, the integral is well-defined modulo a lattice in $\mathbb C$, so the inverse function (your $g(t)$) should be a standard elliptic function, periodic with respect to that lattice.
But for $n\geq 5$, the periods form a dense subset of $\mathbb C$, so the inverse function cannot be well-defined (the level sets of a well-defined holomorphic function are discrete) and in fact fails quite badly, taking infinitely many different values at each point. So this might explain why this function is not well-studied.
A: (This answer doesn't address the question for the whole curve, but really just the question of solving the differential equation near $0$).
I think one way to proceed is as follows, which can be made rigorous and is how things work when $n = 2:$ let $f(t) = \int_{0}^{t} \frac{1}{(1- u^{n})^{1/n}}du$ for $ t \in (-1,1),$ where we take the unique real positive $n$-th root, and let $g$ be the inverse function to $f,$ which exists as $f$ is increasing where defined. Note that $f(0) = 0$ and that $f$ is an odd function if $n$ is even. Note also that (by the Fundamental Theorem of Calculus) $f$ is infinitely differentiable on $(-1,1).$ 
Then $f(g(t)) = t.$ Now $g^{\prime}(t) = \frac{1}{f^{\prime}(g(t))}$ and
$f^{\prime}(g(t)) = \frac{1}{(1-g(t)^{n})^{\frac{1}{n}}}.$ Hence we do have
$g(t)^{n} + g^{\prime}(t)^{n} = 1.$ Note also that $g(0) = 0$ and $g^{\prime}(0) =  1.$ From this, and the equation $g(t)^{n} + g^{\prime}(t)^{n} = 1,$ we can inductively determine the Maclaurin series for $g$.
Later edit partially motivated by Christian Remling's comment. The only difficulty seems to be to find ( near enough $0$) a good choice of $f$ in the complex case, and show that it is injective in  neighbourhood of $0$: For the sake of completeness, I outline a proof that in the complex case, we can construct a solution $g(z)$ with $g(z)^{n} + g^{\prime}(z)^{n} =1,$ and $g(0) = 0,g^{\prime}(0) = 1,$ and that $g(z)$ has a Maclaurin series with strictly positive radius of convergence.
We define $f$ on the open unit disk $C(0;1)$ by $f(z) = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{1+nk}z^{1+nk} \left( \begin{array}{clcr} \frac{-1}{n}\\ k \end{array} \right).$ The power series does converge on $C(0;1),$ and we have 
$f^{\prime}(z) = (1-z^{n})^{\frac{-1}{n}} = \sum_{k=0}^{\infty} (-1)^{k}z^{nk} \left( \begin{array}{clcr} \frac{-1}{n}\\ k \end{array} \right)$ for $z \in C(0;1).$
Let $r$ be a real number with $ 0 < r < (1-\frac{1}{n+1})^{\frac{1}{n}}.$
Then $0 < \frac{r^{n}}{n(1-r^{n})} < 1.$ It is easy to check from the power series that for $z,w \in C(0;r),$ we have 
$|z-w| ( 1 - \frac{r^{n}}{n(1-r^{n})}) \leq  |f(z)-f(w)| \leq |z-w| ( 1 + \frac{r^{n}}{n(1-r^{n})}).$ In particular, $f$ is injective on $C(0;r),$ so the inverse function, $g$ say, is defined on the set $f(C(0;r)),$ which is an open set by the Open Mapping Theorem. The above inequalities imply that $g$ is uniformly continuous on $f(C(0;r)),$ and then it follows that $g^{\prime}(z) = \frac{1}{f^{\prime}(g(z))} = (1-g(z)^{n})^{\frac{1}{n}}.$ Hence we do have $g(z)^{n} + g^{\prime}(z)^{n} = 1$ for $z \in f(C(0;r)).$ Since $g$ is differentiable on an open set which includes $0,$ its Maclaurin series has a strictly positive radius of convergence.
We also note that for any complex $n$-th root of unity, we have $f(\omega z ) = \omega f(z)$ for any $z \in C(0;r)$ and for any (not necessarily primitive) $n$-th root of unity $\omega,$ so that we also have $g(\omega z) = \omega g(z)$ for any such $\omega$ and any $z \in f(C(0;r)).$ It follows easily from this that the Maclaurin series for $g(z)$ has the form $z + \sum_{k=1}^{\infty} a_{k} z^{1+nk}.$
A: Suppose $u,v:\mathbb{C} \to \mathbb{C}$ are entire functions such that $u^{n}+v^{n}=1$ (disregarding relations like $u^{'}=v$ momentarily). Then $\frac{u}{v}$ is a meromorphic function which never attains a value among the $n$ solutions to $z^{n}=−1$. Picard's Little Theorem says that a meromorphic function omitting 3 values from its range is constant. Then $\frac{u}{v}$ is constant so $u$ and $v$ are themselves constant, or $n \leq 2$.
A: There is a general method to reparametrise ANY curve $(x(s),y(s))$ in the form $(g(t),g'(t))$ as required.  We use the simple reparametrisation $$t=\int^s \dfrac {x'(u)}{y(u)} du.$$  In your case, we can employ the parametrisation $(\cos^{2/n}(s),\sin^{2/n}(s))$ as starting point.  
Remarks:


*

*In the general case, one has, of course, to avoid points where the curve crosses the $x$-axis.  In the cases in point here one can restrict attention to the part of the curve which lies in the first quadrant and then use the usual symmetry argument.  One can then use mathematica to attack the resulting integrals (I have no access at the moment).This, of course, leaves open the question of an explicit parametrisation.It will depend on the value of $n$ whether the integral  and the functional inversion required to get $s$ as a function of $t$ can be carried out explicitly.

*I thought that it would be of interest to the OP to see his problem embedded into a more general situation.  For a more careful treatment of how to obtain such a parametrisation for a general curve, see the arXiv paper 1102.1579.  This  also has many examples which show the relevance and usefulness of such parametrisations.
