Expectation of a Geometric Brownian Motion at an independent random time

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Let $W_t$ be a Brownian motion.

I'm interested in the expectation $$ \mathbb{E} \left[ \exp \left\{ \sigma W_\tau \right\} \right] $$ where $\tau$ is a random time independent of $W_t$.

Here's my attempt: $$ \mathbb{E} \left[ \exp \left\{ \sigma W_\tau \right\} \right] = \mathbb{E} \left[ \exp \left\{ \sigma \sqrt{\tau} Z \right\} \right] $$ where $Z$ is a standard normal independent of $\tau$.

Then $$ \begin{aligned} \mathbb{E} \left[ \exp \left\{ \sigma \sqrt{\tau} Z \right\} \right] &= \mathbb{E} \left[ \mathbb{E} \left[ \exp \left\{ \sigma \sqrt{\tau} Z \right\} \big| \tau \right] \right] \\ &= \mathbb{E} \left[ \exp \left\{ \frac{1}{2} \sigma^2 \tau \right\} \right] \\ \end{aligned} $$ which I can then get from the Laplace transform of $\tau$.

Is this argument correct?