Where does one even need Bernoulli's Inequality?

Bernoulli's Inequality can prove the AM-GM Inequality. From this fact, you could derive Young's Inequality, Holder's Inequality, Minkowski's Inequality, and in turn any that follow from those.

Let $a_1, a_2, \ldots, a_n$ be $n$ positive real numbers. Let us define $$A_k=\frac{a_1+a_2+\cdots+a_k}{k}$$ for every $1\leq k\leq n$. Bernoulli's Inequality in the form $x^k\geq 1+k(x-1)$ then implies $$\left(\frac{A_k}{A_{k-1}}\right)^k\geq 1+k\left(\frac{A_k}{A_{k-1}}-1\right)$$ which after some algebraic hyjinx results in $$A_k^k\geq a_kA_{k-1}^{k-1}\,.$$ This in turn implies $$A_n^n\geq a_nA_{n-1}^{n-1}\geq a_na_{n-1}A_{n-2}^{n-2}\geq \cdots\geq a_n\cdots a_2a_1$$ which gives $$\sqrt[n]{a_1\cdots a_n}\leq\frac{a_1+a_2+\cdots+a_n}{n}\,.$$ Intuitively, what's happening here is that we can order the values $A_1, A_2, \ldots, A_n$ so that the subsequent quotients $A_k/A_{k-1}$ are close to $1$, which is where Bernoulli's is precise.

When dealing with asymptotics of probabilities, the expression $(1-p)^n$ comes up all the time. The most convenient way to handle it is with the inequalities $$1 - pn \le (1-p)^n \le e^{-pn}$$ where the lower bound is Bernoulli's inequality. (I'm assuming here that $p \in [0,1]$ and $n$ is a natural number.) Actually, as mentioned in p4sch's answer, the upper bound is also a consequence of Bernoulli's inequality, via the inequality $1 + x \le e^{x}$.

For example, the result that monotone properties of the Erdős–Rényi model of random graphs have thresholds relies on the fact that if you take the union of $k$ copies of $\mathcal G_{n,p}$, the graph you get (which has the distribution $\mathcal G_{n,1-(1-p)^k}$) can be thought of as a subgraph of $\mathcal G_{n,kp}$. This implies that as the edge probability $p$ scales linearly, the probability that your graph lacks a monotone property decays exponentially: $$\Pr[\mathcal G_{n,kp} \text{ lacks property $M$}] \le \Pr[\mathcal G_{n,p} \text{ lacks property $M$}]^k.$$ For more details, see Theorem 1.7 in this textbook.

Many inequalities can prove each other and it's hard to say you ever "need" a particular result. This, however, is an example where Bernoulli's inequality is the most convenient tool to use.

You can use Bernoulli's inequality in order to prove that $\lim_{n \rightarrow \infty} \sqrt[n]{a} =1$, where $a>0$. Here we define the $n$-th square root in elementary fashion by saying it is the solution of $x^n =a$. The existence can be shown by the Babylonian method or simply using statements on the existence of differentiable inverse functions.

Let $x_n +1 = \sqrt[n]{a}$ for (w.l.o.g.) $a \ge 1$. Then $(x_n+1)^n = a \ge 1+nx_n$ and therefore $$\frac{a}{n-1} \ge x_n \ge 0.$$ This proves $x_n \rightarrow 0$. If $a< 1$, then we can apply the previous step with $b= 1/a$ and use that $\sqrt{1/a} = 1/\sqrt{a}$.

Another application: If we define the exponential function via $$\exp(x) := \lim_{n \rightarrow \infty} (1+x/n)^n,$$ then Bernoulli's inequality shows that $$\exp(x) \ge 1+x.$$