What does $o(1)$ stand for here? Let's assume that $ k= o(\sqrt n ) $
In this case,
$$ \sum_{i=1}^{k-1}\ln(1-\frac{i}{n}) \sim  \sum_{i=1}^{k-1} (- \frac{i}{n})= -\frac {1}{n} \frac{(k-1)k}{2}=o(1)
$$
Then, 
$$ e^{o(1)} \sim 1 \quad as \quad n \to \infty
$$
Here I dont understand how this similarities/approximations$\quad(\sim) \quad $are made and also how the equations at third line end up to be $o(1)$
But most importantly I dont know what do we mean by $o(1)$ 
Any definition could you give me? 
 A: $o(1)$ is any function which satisfies a particular property:  It satisfies that $o(1)/1$ goes to $0$ as $n \to \infty$.  In other words, $o(1)$ is just a general name for any expression $f(n)$ which satisfies $f(n)/1 \to 0$ as $n \to \infty$.
Similarly, if I said an expression $g(n)$ is $o(n^{2})$, it means $g(n)/n^{2} \to 0$ as $n \to \infty$.  The interpretation here is that $n^{2}$ grows much faster than $g(n)$ as $n \to \infty$.  How fast?  Fast enough that the ratio $g(n)/n^{2}$ tends to $0$.
So, going back to the original example, if $f(n)$ is $o(1)$, that means $1$ grows much faster than $f(n)$ in the sense that $f(n)/1$ tends to $0$ as $n \to \infty$.  But since $f(n)/1 = f(n)$, that means $f(n)$ tends to $0$.  So, $o(1)$ is notation for any function which tends to $0$ as $n \to \infty$.
To follow up with your question about $e^{o(1)} \sim 1$ as $n \to \infty$, what is meant here is this:  since $o(1)$ is a function of $n$ which goes to $0$ as $n \to \infty$, that means for large enough $n$, the function $o(1)$ is as small as you want (since it keeps tending to $0$).  Ok, but then the function $e^{x}$ is continuous everywhere, so it's continuous at $0$.  That means if $x$ is really close to $0$, then $e^{x}$ is really close to $e^{0} = 1$.  
Now, remember that for large enough $n$, $o(1)$ is really close to $0$.  That means $e^{o(1)}$ is really close to $1$, for large enough $n$ (by the continuity of $e^{x}$ at $x = 0$).  So, $e^{o(1)} \sim 1$ as $n \to \infty$ should be interpreted by you as "for large enough $n$, $e^{o(1)}$ is really close to $1$", and we know this is true since $e^{x}$ is continuous at $x = 0$, and $o(1)$ is close to $0$ for large enough $n$.
A: From the Taylor development
$$\ln\left(1-\frac in\right)\sim-\frac in,$$ and this explains the first similarity.
Then as $k=o(\sqrt n)$, mechanically $\dfrac{k^2}n=o(1)$.
Finally, the antilogarithm of $0$ is $1$.
A: $o(1)$ means that the expression tends towards $0$ when $n\rightarrow \infty$
