Inequality for $\sum_{n=1}^\infty \frac{x^n}{n^n}$.

I need to show that for $x<0$,
$$\sum_{n=1}^\infty \frac{x^n}{n^n}<0$$
but I am completely stuck. I noted that the series is alternating, the first term is negative, but the term is only eventually decreasing. Any hint?

Solutions Collecting From Web of "Inequality for $\sum_{n=1}^\infty \frac{x^n}{n^n}$."

Hint: Notice that $n^n\geqslant n!$ and
$$\sum_{n=1}^\infty \frac{x^n}{n!} = -1+\sum_{n=0}^\infty \frac{x^n}{n!}.$$

Edit: Unfortunately, this hint is problematic when $x<0$. See comment by @FrankS on this post.
I preserve this post because it’s a common way to start thinking.

Edit 2: Comparison of plots actually show that $\sum_{n=1}^\infty \frac{x^n}{n^n} < e^x-1$ when $x<0$. The hint is not useful thus far, but apparently we can get away with this bad math.

Mathematica code for the plots:

Plot[{x Integrate[Power[t, -x t], {t, 0, 1}], Exp[x] - 1}, {x, -3, 3}]

Let $\displaystyle f(x)=\sum_{n=1}^\infty \frac{x^n}{n^n}$ and $\displaystyle g(x)=x\int_0^1 t^{-xt}\,\mathbb dt$.

This method starts from the claim that $f(x)=g(x)$ on $\mathbb R$ to provide a proof for this claim.
Thus when $x<0$, we indeed have $f(x)<0$ because $x$ is negative and $\displaystyle \int_0^1 t^{-xt}\,\mathbb dt$ is positive.

The $n$-th derivative of $g(x)$, written as $g^{(n)}(x)$ is (with the help of Leibniz integral rule)
g^{(n)}(x) = n\int_0^1 (-1)^{n-1}\cdot t^{n-1 – xt}\cdot \ln^{n-1}t\,\mathbb dt
– x\int_0^1 (-1)^{n} \cdot t^{n – xt}\cdot \ln^{n} t\,\mathbb dt.
Therefore, we get $\displaystyle g^{(n)}(0) = n\int_0^1 (-t\ln t)^{n-1}\,\mathbb dt$.

From integration by parts,
n\int_0^1 (-t\ln t)^{n-1}\,\mathbb dt &= (n-1)\int_0^1 t\cdot (-t\ln t)^{n-2}\,\mathbb dt, \\
n\int_0^1 t\cdot (-t\ln t)^{n-2}\,\mathbb dt &= (n-2)\int_0^1 t^2\cdot (-t\ln t)^{n-3}\,\mathbb dt, \quad\cdots \\
we would obtain
g^{(n)}(0) = (n-1)\cdot \frac{n-2}{n}\cdot(\cdots)\cdot\frac{1}{n}\cdot\int_0^1 t^{n-1}\,\mathbb dt = \frac{(n-1)!}{n^{n-1}},
which means that
g(x) = g(0) + \sum_{n=1}^{\infty}\frac{g^{(n)}(0)}{n!}\cdot x^n
= 0 + \sum_{n=1}^{\infty}\frac{(n-1)!}{n^{n-1}\cdot n!} \cdot x^n
= \sum_{n=1}^\infty \frac{x^n}{n^n}.

Therefore, $f$ is identical to $g$; when $x<0$, $\displaystyle \sum_{n=1}^\infty \frac{x^n}{n^n}<0$. $\square$