How does conditional expectation really operate? Let there be a keyboard with k keys, only 9 of which are numbers, which are 1,...,9. A monkey performs a series of random taps. The series will end as the monkey taps a non-number key. Let $N$ be the number of different numbers in the series, and let $X$ be the multiplication of these(those?) different numbers. Let us agree in advance that $X=1$ when $N=0$. Find $E[X|N=n]$. 
The answers say: $E[X|N=n]=E[\prod_{i=1}^{n} X_i]=\prod_{i=1}^{n}E[ X_i]=({1+9\over 2})^n=5^n$. I did go in that direction and did come close, except that I fail to understand how the conditioning part in that expected value computation is taken into consideration. I mean, how do I know it is already taken care of? I only know the technical formula and can't actually discern its fulfillment here. I would appreciate your help here.  

$\mathsf E(X\mid N=n)$ is the (conditional) expected multiple of distinct numbers tapped by the monkey when given that $n$ distinct digit keys are tapped.
Let $X_i$ be the value of the $i$-th distinct key tapped.
Then $\mathsf E(X_i) = \frac {9+1}{2}= 5$ is the expected value of the $i$-th distinct digit.
And so $\mathsf E(\prod_{i=1}^n X_i)$ is the expected multiple of $n$ distinct digits tapped by the monkey.
Hence the expected multiple of distinct numbers tapped by the monkey when $n$ distinct digit keys are tapped is:
$$\begin{align}
\mathsf E(X\mid N=n) & = \mathsf E(\prod_{i=1}^n X_i)
\\ & = \prod_{i=1}^n\mathsf E(X_i)
\\ & = {(\tfrac {10}2)}^{n}
\\ & = 5^n
\end{align}$$
