What is the series representation of $\int_{0}^{t} \exp(-x^n)\, dx$ where $n$ is a positive integer? I have searched in the web to get series representation of $$\int_{0}^{t} e^{-x^n}\,dx$$  where $n$ is a positive integer, but I haven’t succeeded; really I want its representation as hypergeometric series.

Integrating the Power Series
We have the power series
$$
e^{-x^n}=\sum_{k=0}^\infty\frac{(-1)^kx^{kn}}{k!}\tag1
$$
which leads to the alternating power series
$$
\begin{align}
\int_0^t e^{-x^n}\,\mathrm{d}x
&=\sum_{k=0}^\infty\frac{(-1)^kt^{kn+1}}{(kn+1)k!}\\
&=\bbox[5px,border:2px solid #C0A000]{t\,{}_1F_1\!\left(\tfrac1n;1+\tfrac1n;-t^n\right)}\tag2
\end{align}
$$
Mathematica verification:
Simplify[D[t Hypergeometric1F1[1/n,1+1/n,-t^n],t]]

Unilateral Power Series
The problem with $(2)$ is that for large $t$, the terms get huge and oscillate. This means that a lot of significance can get lost in cancellation. It is better to have a unilateral series where nothing is lost to cancellation.
Setting
$$
f(t)=e^{t^n}\int_0^te^{-x^n}\,\mathrm{d}x\tag3
$$
we have
$$
f'(t)=nt^{n-1}f(t)+1\tag4
$$
which, since $f(0)=0$, says not only that
$$
f(t)=t+O\!\left(t^{n+1}\right)\tag5
$$
but also gives the coefficient recurrence
$$
a_{k+n}=\frac{n}{k+n}\,a_k\tag6
$$
From $(5)$ and $(6)$, we get
$$
a_{kn+1}=\frac{\Gamma\!\left(1+\frac1n\right)}{\Gamma\!\left(k+1+\frac1n\right)}\tag7
$$
and $a_k=0$ for $k\not\equiv1\pmod n$. This gives the unilateral series:
$$
\begin{align}
\int_0^te^{-x^n}\,\mathrm{d}x
&=e^{-t^n}\sum_{k=0}^\infty\frac{\Gamma\!\left(1+\frac1n\right)}{\Gamma\!\left(k+1+\frac1n\right)}\,t^{kn+1}\\
&=\bbox[5px,border:2px solid #C0A000]{t\,e^{-t^n}{}_1F_1\!\left(1;1+\tfrac1n;t^n\right)}\tag8
\end{align}
$$
Note that the terms in the sum in $(8)$ peak when $k\sim t^n$.
Mathematica verification:
Simplify[D[t Exp[-t^n]Hypergeometric1F1[1,1+1/n,t^n],t]]

Asymptotic Expansion
$$
\begin{align}
\int_t^\infty e^{-x^n}\,\mathrm{d}x
&=\frac1n\int_{t^n}^\infty e^{-x}x^{\frac1n-1}\,\mathrm{d}x\tag9\\
&=\frac{e^{-t^n}}n\int_0^\infty e^{-x}\left(t^n+x\right)^{\frac1n-1}\mathrm{d}x\tag{10}\\
&=\frac{e^{-t^n}t^{1-n}}n\int_0^\infty e^{-x}\left(1+\frac{x}{t^n}\right)^{\frac1n-1}\mathrm{d}x\tag{11}\\
&=\frac{e^{-t^n}t^{1-n}}n\left[\int_0^{t^n/2}e^{-x}\left(1+\frac{x}{t^n}\right)^{\frac1n-1}\mathrm{d}x+O\!\left(e^{-t^n/2}\right)\right]\tag{12}\\
&=\frac{e^{-t^n}t^{1-n}}n\left[\sum_{k=0}^{m-1}(-1)^k\int_0^{t^n/2} e^{-x}\frac{\Gamma\left(k+1-\frac1n\right)}{k!\,\Gamma\left(1-\frac1n\right)}\frac{x^k}{t^{nk}}\mathrm{d}x+O\!\left(\frac1{t^{nm}}\right)\tag{13}\right]\\
&=\frac{e^{-t^n}t^{1-n}}n\left[\sum_{k=0}^{m-1}(-1)^k\frac{\Gamma\left(k+1-\frac1n\right)}{\Gamma\left(1-\frac1n\right)}\frac1{t^{nk}}+O\!\left(\frac1{t^{nm}}\right)\right]\tag{14}
\end{align}
$$
Explanation:
$\phantom{0}(9)$: substitute $x\mapsto x^{1/n}$
$(10)$: substitute $x\mapsto t^n+x$
$(11)$: substitute $x\mapsto x/t^n$
$(12)$: $\int_{t^n/2}^\infty e^{-x}\,\mathrm{d}x=e^{-t^n/2}$
$(13)$: Binomial Theorem
$(14)$: $\int_{t^n/2}^\infty e^{-x}x^k\,\mathrm{d}t=O\!\left(e^{-t^n/3}\right)$
Therefore,
$$
\begin{align}
\int_0^te^{-x^n}\,\mathrm{d}x
&=\Gamma\left(1+\frac1n\right)-\frac{e^{-t^n}t^{1-n}}n\left[\sum_{k=0}^{m-1}(-1)^k\frac{\Gamma\left(k+1-\frac1n\right)}{\Gamma\left(1-\frac1n\right)}\frac1{t^{nk}}+O\!\left(\frac1{t^{nm}}\right)\right]\\
&=\bbox[5px,border:2px solid #C0A000]{\Gamma\left(1+\frac1n\right)-\frac{e^{-t^n}t^{1-n}}n\,{}_2F_0\!\left(1,1-\tfrac1n;;-t^{-n}\right)}\tag{15}
\end{align}
$$
where ${}_2F_0\!\left(1,1-\tfrac1n;;-t^{-n}\right)$ is not convergent, but is to be taken in the asymptotic sense.
