My professor said that, assuming independence, likelihood functions are basically crafted from the product of the individual probability density functions for each observation in a sample.
Thus, for example, if we are talking about the exponential distribution, then the likelihood function will look something like \begin{align} L(\theta; x_1, \dots, x_n) & = \prod^n_{i = 1} f_X(x_i; \theta)\\ & = \prod^n_{i = 1} \frac 1 {\theta} e^{-\frac {x_i} {\theta}}. \end{align}
If I follow this logic, then why is it, for a binomial distribution, the likelihood function is not \begin{align} L(p; x_1, \dots, x_n) & = \prod^n_{i = 1} f_X(x_i; p)\\ & = \prod^n_{i = 1} \binom n {x_i} p^{x_i} (1 - p)^{n - x_i} \end{align} but rather $$L(p; x_1, \dots, x_n) = \binom n x p^x (1 - p)^{n - x}?$$ Is the latter not simply the probability mass function of the binomial distribution? Why do we not need to take the product in this case? What am I missing here?
I am still finding the concept a little confusing, so any help will be greatly appreciated :)
This equation $$L(p; x_1, \dots, x_n) = \binom n x p^x (1 - p)^{n - x} \tag{1a}$$ that you wrote does not make sense. The LHS contains the observed data $(x_1, \ldots, x_n)$, but on the RHS, you have just $x$, which you do not define.
Moreover, in both of your equations, the $n$ on the LHS pertains to the sample size (the number of observations), whereas on the RHS, $n$ corresponds to the number of trials in the binomial distribution. So $$\begin{align} L(p; x_1, \dots, x_n) & = \prod^n_{i = 1} f_X(x_i; p)\\ & = \prod^n_{i = 1} \binom n {x_i} p^{x_i} (1 - p)^{n - x_i} \end{align} \tag{1b}$$ is also incorrect, unless you mean that the number of trials in your binomial distribution always equals your sample size for the likelihood.
That said, if $x_1, \ldots, x_n$ are independent and identically distributed observations from a Bernoulli distribution with $\Pr[X_i = 1] = p$, then the sum $$X = \sum_{i=1}^n X_i$$ is binomial with parameters $n$ and $p$, and the likelihood of $p$ is $$\mathcal L(p; x_1, \ldots, x_n) \propto \prod_{i=1}^n p^{x_i} (1-p)^{1 - x_i} = p^x (1-p)^{n-x}. \tag{2}$$ Note the binomial coefficient $\binom{n}{x}$ is not necessary because it is a constant with respect to $p$.
In the general case, if $x_1, \ldots, x_n$ are binomial with parameters $m$ and $p$ (note the use of $m$ because $n$ is being used to describe the number of binomial observations), then the joint likelihood of $p$ is $$\mathcal L(m, p; x_1, \ldots, x_n) = \prod_{i=1}^n \binom{m}{x_i} p^{x_i} (1-p)^{m - x_i} = \left(\prod_{i=1}^n \binom{m}{x_i}\right) p^{n \bar x} (1-p)^{n(m - \bar x)},\tag{3}$$ where $$\bar x = \frac{1}{n} \sum_{i=1}^n x_i. \tag{4}$$ If $m$ is fixed and known, then the likelihood of $p$ can ignore the product of binomial coefficients and we can write it as $$\mathcal L(p; m, x_1, \ldots, x_n) = p^{n \bar x} (1 - p)^{n(m-\bar x)}. \tag{5}$$ Then when $m = 1$, we recover $(2)$ from $(5)$.