Identifying a power series I'm no analyst, so when a student in the class to whom I was teaching some elementary theory of (power) series, asked about this:
$\displaystyle{\sum_{n=0}^\infty\frac{x^n}{n^n}=1+x+\left(\frac{x}{2}\right)^2+\left(\frac{x}{3}\right)^3+\left(\frac{x}{4}\right)^4+\cdots}$
(assuming $0^0=1$), I had no idea.  It didn't look like anything I recognized, and a play about with some derivatives gave me no useful information.
Is this a known function, and does this series admit of an explicit formula?
 A: To start analyzing this series, you could consider using Sterling's approximation $n!\approx \left(\frac ne\right)^n\sqrt {2\pi n}$:
$$\sum_{n=0}^\infty\left(\frac xn\right)^n\approx \sqrt{2\pi}\sum_{n=0}^\infty \frac {\sqrt n({x\over e})^n}{n!}\ge \sqrt{2\pi}\sum_{n=0}^\infty \frac {({x\over e})^n}{n!}=e^{\frac xe} \sqrt{2\pi}$$
Of course this is only a lower limit to the approximated value, and that $\sqrt n$ term still needs to be dealt with...
A: I had to do a lot of research on this, which was my benefit :)
Outline. Basically, I will use the Sophomore's Dream to deduce an integral expression for the answer.
Lemma. We see
$$\int^{1}_{0}x^{n}\ln(x)^{n}\,\mathrm{d}x = (-1)^{n}(n+1)^{-(n+1)}n!\tag{1}$$
Proof. We change variables writing
$$\tag{2a}x=\exp\bigl(-u/(n+1)\bigr)$$
which lets us rewrite (1) as
$$\int^{1}_{0}x^{n}\ln(x)^{n}\,\mathrm{d}x = (-1)^{n}(n+1)^{-(n+1)}\int^{\infty}_{0}u^{n}e^{-u}\,\mathrm{d}u\tag{2b}$$
Observe the integral on the right hand side is precisely $n!$ (thanks to the Gamma function). And that concludes the proof for our lemma. QED.
Theorem. We claim
$$f(t) = 1+\sum^{\infty}_{n=1}\left(\frac{t}{n}\right)^{n} = 1 + t\int^{1}_{0}x^{-xt}\,\mathrm{d}x.\tag{3}$$
Proof. We end up rewriting the integrand on the right hand side
$$x^{-xt} = e^{-xt\ln(x)} = \sum^{\infty}_{n=0}\frac{(-t)^{n}}{n!}x^{n}\ln(x)^{n}.\tag{4a}$$
We plug this back into the integral
$$\int^{1}_{0}x^{-xt}\,\mathrm{d}x= \int^{1}_{0}\sum^{\infty}_{n=0}\frac{(-t)^{n}}{n!}x^{n}\ln(x)^{n}\,\mathrm{d}x.\tag{4b}$$
Swap the sum and integral
$$\int^{1}_{0}\sum^{\infty}_{n=0}\frac{(-t)^{n}}{n!}x^{n}\ln(x)^{n}\,\mathrm{d}x
=\sum^{\infty}_{n=0}\frac{(-t)^{n}}{n!}\int^{1}_{0}x^{n}\ln(x)^{n}\,\mathrm{d}x.\tag{4c}$$
We use our lemma to rewrite the right hand side as
\begin{align}
\sum^{\infty}_{n=0}\frac{(-t)^{n}}{n!}\int^{1}_{0}x^{n}\ln(x)^{n}\,\mathrm{d}x &= \sum^{\infty}_{n=0}\frac{(-t)^{n}}{n!}\left((-1)^{n}(n+1)^{-n+1}n!\right)\\
&= \sum^{\infty}_{n=0}\frac{t^{n}}{(n+1)^{n+1}}.\tag{4c}
\end{align}
Then we just fiddle with arithmetic (multiply by $t$, and add 1) to get the series in question. That concludes our proof for the theorem. QED.
Remark. There is no closed form expression for the integral that I am aware of. Perhaps either the OP or some other user knows some fancy-pants way to evaluate the integral, but I do not know of one readily available :(
