Showing $\frac{1}{2\pi}\frac{\exp(i(v-w))t}{i(v-w)} |^{a+2\pi}_{a} = 0$, for $v \neq w$

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I am trying to prove an exponential identity and am stuck at:

$$\frac{1}{2\pi}\frac{\exp(i(v-w))t}{i(v-w)} |^{a+2\pi}_{a} = 0, \quad \text{for $v \neq w$}$$

The textbook suggests that this is equal because $\exp$ is $2\pi$ periodic which seems clear to me. But wouldn't it also be possible to rewrite this as:

$$\frac{1}{2\pi}\frac{\exp(i(v-w))2\pi}{i(v-w)} = \frac{\exp(i(v-w))}{i(v-w)}$$ ?