$n$-th power of stochastic exponential

The following is true: if $X$ does not jump by -1 and $\alpha\neq 0$ is an arbitrary complex number, then $$\mathcal{E}(X)^\alpha = \mathcal{E}(Y), $$ where $$Y = ((1+\mathrm{id})^\alpha-1)\circ X.$$ Here $f\circ X$ denotes the $f$-variation of $X$.

The function $f = (1+\mathrm{id})^\alpha-1$ is analytic at zero, hence for $X$ continuous one has by Émery's formula $$ Y = f'(0)\cdot X + \frac{1}{2}f''(0)\cdot [X,X].$$ For details see https://arxiv.org/abs/2006.11914.

You can also do this by hand, write down the Ito formula for $Z=\mathcal{E}(X)^\alpha$ and observe that it gives $dZ_t = Z_t dY_t$ with $Z_0=1$. Hence $Z=\mathcal{E}(Y)$.

Recall that for a continuous semimartingale $Z_t$ that starts at zero, its Doléans-Dade exponential is given by $$\mathcal{E}(Z)_t = \exp \left( Z_t - \frac{1}{2}\left\langle Z \right\rangle_t \right)$$

Thus, for $X$, we have $$\mathcal{E}(X)_t^n = \exp \left( nX_t - \frac{1}{2}n\left\langle X\right\rangle_t \right)$$

Now, note that $\left\langle Y \right\rangle_t = \left\langle nX \right\rangle_t = n^2\left\langle X \right\rangle_t$, since the second term in the definition of $Y$ is of locally bounded variation. Thus, $$ \begin{align*} \mathcal{E}(Y)_t &= \exp \left( nX_t + \frac{1}{2}n(n-1)\left\langle X \right\rangle_t - \frac{1}{2}n^2\left\langle X \right\rangle_t\right) \\ &= \exp \left( nX_t - \frac{1}{2}n\left\langle X \right\rangle_t\right) \\ &= \mathcal{E}(X)_t^n \end{align*}$$ As desired.