Other ways to show $y_n = \prod_{k=1}^n \left(1+{1\over x_k}\right)$ is bounded, where $x_n$ is another sequence with given properties. Problem statement:

Let $n\in \mathbb N$ and $\{x_n\}$ be a sequence of natural numbers such that $S_n$ defined by:
  $$
S_n = {1\over x_1} + {1\over x_2} + {1\over x_2} + \dots + {1\over x_n}
$$
  is bounded. 
Show that $y_n$ is bounded given:
  $$
y_n = \prod_{k=1}^n\left(1 + {1\over x_k}\right)
$$

From the problem statement we know that $S_n$ is bounded. Let's try to take log of $y_n$:
$$
\ln y_n = \ln\left(\prod_{k=1}^n\left(1 + {1\over x_k}\right)\right) = \ln\left(1 + {1\over x_1}\right) + \ln\left(1 + {1\over x_2}\right) + \dots +\ln\left(1 + {1\over x_n}\right)
$$
We know that $x_k$ is a natural number and therefore $x_k > -1$. Based on that we may use the following inequality:
$$
\ln(1+x) < x, \; \forall x>-1
$$
Thus:
$$
\ln y_n = \sum_{k=1}^n\left(\ln\left(1+{1\over x_k}\right)\right) <\sum_{k=1}^n{1\over x_k} = S_n
$$
But we know that $S_n$ is bounded, now given $S_n < M$:
$$
\ln y_n <M \iff y_n <e^M
$$
Thus $e^M$ is some constant being an upper bound for $y_n$ and $y_n > 0$. So finally:
$$
0<y_n < e^M
$$
Question:

I would like to know whether my proof is valid and find(if possible) a precalculus way of proving this, since $\ln(1+x) < x$ requires derivatives to be proven.

 A: The OP's solution is fine, but it is technically not a precalculus solution.  Here is a solution without even using logarithm.  I only require that $x_1,x_2,\ldots$ are positive real numbers such that $S_n$ converges.  (Well, I could also allow negative $x_k$ as long as $\sum_{k=1}^\infty\frac{1}{x_k}$ converges absolutely, but because the OP wants an elementary method, this more general assumption is not being made.)  
Since $S_n$ is bounded, there can be only finitely many $k$ with $0<x_k<1$.  We claim that, for some $m$, $$\sum_{k=m+1}^n\frac{1}{x_k}<\frac12$$
for every $n>m$.  Suppose contrary that, for each $m$, there exists $n_m>m$ for which
$$\sum_{k=m+1}^{n_m}\frac{1}{x_k}\geq \frac12.$$
Set $t_0=0$, $t_1=n_{t_0}$, $t_2=n_{t_1}$, $t_3=n_{t_2}$, $\ldots$.  Then,
$$\sum_{k=t_j+1}^{t_{j+1}}\frac{1}{x_k}\geq \frac12$$
for $j=0,1,2,\ldots$.  In particular,
$$S_{t_l}=\sum_{k=1}^{t_l}\frac{1}{x_k}=\sum_{j=0}^{l-1}\sum_{k=t_j+1}^{t_{j+1}}\frac{1}{x_k}\geq \sum_{j=0}^{l-1}\frac12\geq \frac{l}{2}$$
for all $l=1,2,3,\ldots$.  This contradicts the assumption that $S_n$ is bounded.
So, there does exist $m$ such that $\sum_{k=m+1}^n\frac{1}{x_k}<\frac12$ for all $n>m$.  Thus,
$$y_n=\prod_{k=1}^m\left(1+\frac{1}{x_k}\right)\prod_{k=m+1}^n\left(1+\frac{1}{x_k}\right)$$
for $n>m$.  Now,
$$\prod_{k=m+1}^n\left(1+\frac{1}{x_k}\right)=\frac{\prod_{k=m+1}^n\left(1-\frac{1}{x_k^2}\right)}{\prod_{k=m+1}^n\left(1-\frac{1}{x_k}\right)}<\frac{1}{\prod_{k=m+1}^n\left(1-\frac{1}{x_k}\right)}.$$
By induction,
$$\sum_{k=m+1}^n\left(1-\frac{1}{x_k}\right)\geq 1-\sum_{k=m+1}^n\frac{1}{x_k}>\frac12.$$
Hence, for $n>m$, 
$$y_n<2\prod_{k=1}^m\left(1+\frac{1}{x_k}\right)<\infty.$$
Because $m$ is fixed, the sequence $\left(y_n\right)$ is indeed bounded.
