Showing $ \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^q \to E((X-EX)^q)$ in probability [closed]

Question

I am trying to show for $X_n$ iid st. $E|X|^q < \infty$ that

$$ \frac{1}{n} \sum^n (X_i - \bar{X})^q \to E((X-EX)^q) $$ in probability.

I used the following approach: We note

• $ \frac{1}{n} \sum_{i=1}^n X_i \to EX$ in prob. by the WLLN, i.e., $\bar{X}_n \to EX$ in probability
• $ Y_i := (X_i - EX)^q $ are iid

i.e. we use Slutsky on $\frac{1}{n} \sum^n (X_i - \bar{X})^q $ to obtain iid random variables to then use WLLN as per below:

$$ \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^q \quad \xrightarrow{\text{Slutsky}} \quad \frac{1}{n} \sum^n (X_i - EX)^q \quad\xrightarrow{\text{WLLN}} \quad E(X-EX)^q \>. $$

Now, I realised that, of course, once I applied Slutsky I have taken $ n \to \infty $ so I cant really take the last step via the WLLN. Can somebody help me fix this proof ?