When trying to prove, in this previous question, that $\sum_{n=1}^\infty \frac{x^n}{n^n}<0$ when $x<0$, a common fallacy is incited by the resemblance between the given summation and $\sum_{n=1}^\infty \frac{x^n}{n!}$:
Because $n^n\geqslant n!$ for $n\geqslant 1$,
$$\sum_{n=1}^\infty \frac{x^n}{n^n} \leqslant \sum_{n=1}^\infty\frac{x^n}{n!}=\left(\sum_{n=0}^\infty\frac{x^n}{n!}\right)-1=e^x-1.$$
Unfortunately, this proof only works for the case where $x\geqslant 0$, because the bounding on the summand doesn't hold for $x<0$ when the power $n$ is odd.
However, this invalid proof doesn't imply that $\sum_{n=1}^\infty \frac{x^n}{n^n} \leqslant e^x-1$ doesn't hold for $x<0$. As a matter of fact, we could observe the following figure, where the lower, $\color{blue}{\text{blue}}$ line is $$\sum_{n=1}^\infty \frac{x^n}{n^n}=x\int_0^1 t^{-xt}\,\mathbb dt,$$ and the higher, $\color{purple}{\text{purple}}$ line is $e^x-1$.
Therefore, how to prove that $\displaystyle\sum_{n=1}^\infty \frac{x^n}{n^n} \leqslant e^x-1$ for all $x\in\mathbb R$? (Equality holds at and only at $x=0$.)
Let us set $0^0=1$ just to simplify the notation.We have to prove that for any $x>0$ $$ \sum_{n\geq 0}\frac{(-1)^n}{n^n} x^n \leq e^{-x} \tag{1}$$ holds, or that $$ \sum_{n\geq 0}\frac{(-1)^n}{n^n}x^n\sum_{m\geq 0}\frac{1}{m!}x^m \leq 1 \tag{2}$$ holds. By writing the LHS of $(2)$ as a Cauchy convolution, we need to prove
$$ \sum_{l\geq 0}\left(\sum_{k=0}^{l}\frac{(-1)^k}{k^k (l-k)!}\right) x^l \leq 1 \tag{3}$$ or: $$ \sum_{l\geq 1}\left(\frac{1}{l!}+\sum_{k=1}^{l}\frac{(-1)^k}{k^k(l-k)!}\right)x^l\leq 0\tag{4}$$ and we may check by induction that the involved coefficients are all $\leq 0$.