What is exact value of $\sum_{n=1}^{\infty} \frac{1}{n^n}$ . [duplicate]

This question already has an answer here:

I would like to ask if anyone would help me with solving the following infinite series. $$\begin{equation} \sum_{n=1}^{\infty} \frac{1}{n^n} = \,? \end{equation}$$ Thank you in advance.

marked as duplicate by Hans Lundmark, metamorphy, Xander Henderson, Daryl, José Carlos Santos sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 7 at 22:12

Although it has no closed form, it does satisfy the remarkable identity $$\sum_{n=1}^\infty \frac{1}{n^n} = \int_0^1\frac{dx}{x^x}.$$ It is also equal to $$\ln(3)^e$$ to five decimal places.