Value of an exponential function What is the value of the exponential function $y=e^{x^{x}-1}$ at $x=0$? I graphed the function on desmos and the value at $x=0$ is $1$. However, I do not know how to show this analytically.

Sometimes we express $0^0 = 1$, then we have $e^{0^0-1}=e^{1-1}=e^0=1$. However, normally $x^x$ is undefined at $x=0$. So we take the limit, $ \lim_{x\to 0} (e^{x^x-1}) $. Substitute $x \mapsto x^x$ (see below),
$$ \lim_{x\to 1} (e^{x-1}) = e^0 = 1 $$
Here I show that $x^x \to 1$ as $x \to 0$
\begin{align}
y &= \lim_{x \to 0} x^x \\
\ln y &= \lim_{x \to 0} x\ln x \\
&= \lim_{x\to 0} \frac{\ln x}{\frac 1 x} \\
&= \lim_{x \to 0} \frac{ \frac 1 x}{\frac{-1}{x^2}} \\
\ln y & = \lim_{x \to 0} (-x) = 0 \\
y &= 1
\end{align}
