$e^{a 2\pi i} = (e^{2\pi i})^a$. When $a$ is any real number , Is it true $e^{a 2\pi i} = (e^{2\pi i})^a$ ?
The reason why I ask this question is that I met this situation wheter this equality hold in Calculating Integral in Complex variables.
 A: In general, we define $b^a$ for $b \ne 0$ as $\exp(a \log b)$, but the logarithm is multivalued: if $\text{Log}(b)$ is one of its values, the others are $\text{Log}(b) + 2 \pi i n$ for integers $n$.  Of course $\exp(2\pi i) = 1$.  We can take $\text{Log}(1) = 0$, so the values of 
$(\exp(2\pi i))^a$ are $\exp(2\pi i a n)$ for integers $n$.  In particular $\exp(2\pi i a)$ is always one of the values, but not the only one unless $a$ is an integer.  
The principal branch of $\log(b)$ is the value with imaginary part in $(-\pi ,\pi]$, and the principal branch of $b^a$ is the value of $\exp(a \log(b))$ using the principal branch of $\log(b)$.  If you were to pick out a single value for these multivalued functions, this might be the most popular choice.  The principal branch of $\log(1)$ is $0$.  So for the principal branch, $1^a = 1$, which is not $\exp(2\pi i a)$ unless $a$ is an integer.
A: No, it is not always true. The issue is that exponentiating can sometimes lead to multivaluedness. Here is an example with $a=\frac{1}{2}$,
$\begin{equation*}
\left(e^{2\pi i}\right)^a=\left(1\right)^{\frac{1}{2}}=\pm 1\neq -1=e^{\pi i}=e^{a2\pi i}
\end{equation*}$
