Another way of motivating the use of exponential function when deriving Chernoff bounds from Markov inequality is by considering sums of random variables. The reason is that we can turn a sum into a product, as $e^{X_1 + X_2} = e^{X_1} e^{X_2}.$ Furthermore, if $X_1,$ $X_2$ are independent, then $\mathbb{E}[e^{X_1} e^{X_2}] = \mathbb{E}[e^{X_1}] \mathbb{E}[e^{X_2}]$, and the same would hold if we multiplied each $X_i$ by $\lambda > 0$.
Now consider $n$ i.i.d. random variables $X_1, X_2, \dots, X_n$ that take values +1 and -1, each with probability $\frac{1}{2}.$ These random variable are known as Rademacher random variables. They are convenient for the derivation I'll do below. It's possible to do a similar derivation for other bounded random variables, but the calculation is more annoying.
What we will do is apply Markov inequality to $e^{\lambda S_n},$ but to do so, we need to bound the moment generating function $\mathbb{E}[e^{\lambda S_n}].$ Based on our observation above and because $X_i$'s are i.i.d., we have
\begin{align*}
\mathbb{E}[e^{\lambda S_n}] &= \mathbb{E}[e^{\lambda (X_1 + X_2 + \dots + X_n)}]\\
&= \mathbb{E}[e^{\lambda X_1} e^{\lambda X_2}\dots e^{\lambda X_n}]\\
&= \mathbb{E}[e^{\lambda X_1}]\mathbb{E}[e^{\lambda X_2}]\dots \mathbb{E}[e^{\lambda X_n}]\\
&= \Big(\mathbb{E}[e^{\lambda X_1}]\Big)^n.
\end{align*}
Now, $X_1$ takes values +1 and -1 with probability $\frac{1}{2}$ each so by explicitly writing the expectation above we have $\mathbb{E}[e^{\lambda X_1}] = \frac{1}{2}e^{\lambda} + \frac{1}{2}e^{-\lambda}.$ Recall now once again that the Taylor series for $e^{\lambda}$ is $1 + \frac{\lambda}{1!} + \frac{\lambda^2}{2!} + \frac{\lambda^3}{3!}\dots,$ and, similarly, $e^{-\lambda} = 1 - \frac{\lambda}{1!} + \frac{\lambda^2}{2!} - \frac{\lambda^3}{3!}\dots.$ Hence, we obtain the bound
\begin{align*}
\mathbb{E}[e^{\lambda S_n}] &= \Big(1 + \frac{\lambda^2}{2!} + \frac{\lambda^4}{4!} + \dots\Big)^n\\
&\leq \Big(1 + \frac{\lambda^2}{2 \cdot 1!} + \frac{\lambda^4}{2^2 \cdot 2!} + \dots\Big)^n\\
&= \big(e^{\lambda^2/2}\big)^n\\
&= e^{\frac{\lambda^2 n}{2}}.
\end{align*}
Applying Markov inequality to $e^{\lambda S_n}$ and combining with the bound on $\mathbb{E}[e^{\lambda S_n}]$ we have just derived, we get
$$
\mathbb{P}[S_n \geq t] \leq \frac{\mathbb{E}[e^{\lambda S_n}]}{e^{\lambda t}} \leq e^{\lambda^2 n/2 - \lambda t}.
$$
Minimizing the right-hand side over $\lambda$ (and observe it's essentially the same function as we had for Gaussian random variables), we get that by setting $\lambda = \frac{t}{n},$
$$
\mathbb{P}[S_n \geq t] \leq e^{-\frac{t^2}{2n}}.
$$
Hence, we got a bound that looks suspiciously similar to the bound for Gaussian random variables we derived before. In fact, it is the same bound if we set $\sigma^2 = n.$ This is not a coincidence, as we see in what follows.
Before moving on, notice that I could have likewise chosen to bound the tails of $\frac{S_n}{n}.$ It should be immediate (if not, repeat the derivation above as an exercise!) to you that
$$
\mathbb{P}[S_n/n \geq t] \leq e^{-\frac{t^2 n}{2}}.
$$
