Answer
First step
$e^{S_{n}} = exp (1 + \frac{1}{2} +\frac{1}{3} +\frac{1}{4} +...+ \frac{1}{n}) = e^{1} \times e^{\frac{1}{2}} \times e^{\frac{1}{3}} \times e^{\frac{1}{4}} ... e^{\frac{1}{n}}$
The rest is shown below to prove that the harmonic series is divergent.
Work Step by Step
$e^{S_{n}} = exp (1 + \frac{1}{2} +\frac{1}{3} +\frac{1}{4} +...+ \frac{1}{n}) = e^{1} \times e^{\frac{1}{2}} \times e^{\frac{1}{3}} \times e^{\frac{1}{4}} ... e^{\frac{1}{n}}$
Since $e^{n} \gt 1+n$
$e^{1} \times e^{\frac{1}{2}} \times e^{\frac{1}{3}} \times e^{\frac{1}{4}} ... e^{\frac{1}{n}} \gt (1+1) \times (1+\frac{1}{2}) \times (1+\frac{1}{3}) \times (1+\frac{1}{4}) \times ... \times (1+\frac{1}{n})$
$e^{S_{n}} \gt (\frac{2}{1}) \times (\frac{3}{2}) \times (\frac{4}{3}) \times (\frac{5}{4}) \times ... \times (\frac{n+1}{n})$
$e^{S_{n}} \gt n+1$
$\ln e^{S_{n}} \gt \ln(n+1)$
$S_{n} \gt \ln (n+1)$
$\lim\limits_{n \to \infty}S_{n} \gt \lim\limits_{n \to \infty} \ln(n+1)$
Sum of harmonic series $\gt \infty$
Hence the harmonic series diverges.
