Suppose the prior for $\mu$ is $\sim gamma(k, \lambda)$, so
$f_{\mu}(m) = C m^{k - 1} \exp(-\lambda m)$ when $(m \geq 0)$
Further, suppose the data $X_1, \ldots, X_n$ given $\mu$ is iid Poisson with parameter $\mu$, so for each $[X_j \mid \mu = m] \sim [X \mid \mu = m]$,
$f_{X \mid \mu = m}(i) = \exp(-m) m^i / i!$ when $(i \in \{0, 1, \ldots\})$ Show that the posterior for $\mu$ given the data is also Gamma.
I know that the posterior = prior * Likelihood but am not sure how to recognize the Gamma.
First, observe that the posterior is not exactly what you wrote but it is $\propto$, say it is prior $\times$ L $\times$ constant
Second, your L is the product of your $f$
$$L= C e^{-m n}m ^{\Sigma_k i_k}$$
In the constant C are included all the quantities that do not depend on $m$
Thus the posterior is
$$C m^{(k+\Sigma_k i_k)-1}e^{-(\lambda+n)m}$$
Here the variable is $m \geq 0$ and the posterior is in the same type of your prior, thus it is still gamma