Why is the probability of drawing a certain element from a population of size $N$ be $1/N$ even though the sampling is done without replacement? Related:
Trying to understand the derivation of expectation of sample mean $E(\bar x)= \mu$ where $\mu$ is the mean of the population

I've been trying to understand $\mathrm E(x_i)= \mu$ where $x_i$ is the element at the $i^{\textrm {th}}$ draw from a population of size $N$ for making a sample of size $n$ by SRSWOR.
I read between the lines of my book again; here is the excerpt:

Let $x_1,x_2,\ldots,x_n$ represent a simple random sample from the population $X_1,X_2,\ldots,X_n$. In SRSWR, the probability of selection of any particular member of the population at any drawing remains a constant $1/N$; because before any drawing the population contains all the $N$ members. It may be shown that this result is also true in SRSWOR, although the population size varies at each stage of the selection. Thus, the probability of obtaining the population member $X_k$(suppose) at the $i$-th drawing is a constant $1/N$ both in SRSWR and SRSWOR; i.e., $$\mathrm P(x_i= X_k)= \frac{1}{N}$$

While I'm sure of this for SRSWR, I'm not getting this for SRSWOR.
How can the probability of drawing a certain element from the population at a certain draw be $1/N$ for even the author says 'the population size varies at each stage of the selection'?
If after each turn, the population size is not the same, hoe can still the probability of drawing certain element be $1/N$ when the sampling is done without replacement?
Can anyone please explain me the case for SRSWOR? 
 A: By way of induction and the total law of probability it holds that:
$$P(X_2 = k) = \sum_{i=1}^n P(X_2 = k | X_1 = i) P(X_1 = i)$$
and supposing it holds for any $j$ that $P(X_j = k) = 1/N$ then 
$$P(X_{j+1} = k) = \sum_{i=1} ^ N P(X_{j+1} = k | X_j = i) P(X_j = i)$$.
I think the comments I've contributed on the post contribute more of the intuition behind this finding.
A: Let $A_l^{\textrm {j}}$ denotes an event that a particular unit $X_j$ is not selected at the $l^{\textrm {th}}$ draw. The probability of selecting, say, $k^{\textrm {th}}$ unit at $i^{\textrm {th}}$ draw is: (taking $A_i^{\textrm {k}}$= $A_i$)
\begin{align}
P (x_i = X_k) &= P(A_1 \cap A_2 \cap A_3 \cap A_4 .... A_{i-1} \cap \overline{A_i})\\
              &= P(A_1)P(A_2|A_1)P(A_3|A_1A_2).....P(\overline{A_i}|A_1A_2...A_{i-1}) \\
              &= (1-\frac{1}{N})(1-\frac{1}{N-1})...(1-\frac{1}{N-i+2})\frac{1}{N-i+1}\\
              &= (\frac{N-1}{N})(\frac{N-2}{N-1})...(\frac{N-i+1}{N-i+2})\frac{1}{N-i+1}\\
              &= \frac{1}{N}
\end{align}
