Is there an elementary proof that $\sum \limits_{k=1}^n \frac1k$ is never an integer? If $n>1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.
If you know Bertrand's Postulate, then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.
Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong.

An elementary proof uses the following fact:
If $2^s$ is the highest power of $2$ in the set $S = \{1,2,...,n\}$, then $2^s$ is not a divisor of any other integer in $S$.
To use that, 
consider the highest power of $2$ which divides $n!$. Say that is $t$.
Now the number can be rewritten as 
$\displaystyle \frac{\sum \limits_{k=1}^{n}{\frac{n!}{k}}}{n!}$
The highest power of $2$ which divides the denominator is $t$.
Now the highest power of $2$ that divides $\displaystyle \frac{n!}{k}$ is at least $t-s$. If $k \neq 2^{s}$, then this is atleast $t-s+1$ as the highest power of $2$ that divides $k$ is atmost $s-1$.
In case $k=2^s$, the highest power of $2$ that divides $ \dfrac{n!}{k}$ is exactly $t-s$.
Thus the highest power of $2$ that divides the numerator is atmost $t-s$. If $s \gt 0$ (which is true if $n \gt 1$), we are done. 
In fact the above proof shows that the number is of the form $\frac{\text{odd}}{\text{even}}$.
