I already answered part (a). It was straight forward that the answer to that is $$f_{X_1...X_n}(x_1...x_n)=f_{X_1}(x_1)f_{X_2}(x_2)...f_{X_n}(x_n)$$ assuming mutual independence.
Unfortunately, it is part (b) where I become confused. I know that it holds true because we have learned about the properties of expectations when it comes to independent random variables previously, but how can I $show$ it. In other words, can somebody assist me in breaking it down. I always want to make sure that I have my fundamentals in tact before proceeding to more complicated problems.

According to the law of the unconscious statistician and based on the fact the r.v. are iid, we have that \begin{align*} \textbf{E}(f_{X}(x_{1},x_{2},\ldots,x_{n})) & = \int_{\mathbb{R}^{n}}f_{X}(x_{1},x_{2},\ldots,x_{n})f_{X}(x_{1},x_{2},\ldots,x_{n})\mathrm{d}x_{1}\ldots\mathrm{d}x_{n}\\\\\ & = \int_{\mathbb{R}^{n}}[f_{X_{1}}(x_{1})\ldots f_{X_{n}}(x_{n})][f_{X_{1}}(x_{1})\ldots f_{X_{n}}(x_{n})]\mathrm{d}x_{1}\ldots\mathrm{d}x_{n}\\\\ & = \int_{\mathbb{R}}[f_{X_{1}}(x_{1})]^{2}\mathrm{d}x_{1}\int_{\mathbb{R}}[f_{X_{2}}(x_{2})]^{2}\mathrm{d}x_{2}\times\ldots\times\int_{\mathbb{R}}[f_{X_{n}}(x_{n})]^{2}\mathrm{d}x_{n}\\\\ & = \textbf{E}(f_{X_{1}}(x_{1}))\times\textbf{E}(f_{X_{2}}(x_{2}))\times\ldots\times\textbf{E}(f_{X_{n}}(x_{n}))\\\\ & = [\textbf{E}(f_{X_{1}}(x_{1}))]^{n} \end{align*}
and we are done.
Hopefully this helps!