How $\prod_{j=k+1}^{n}(e^{-\lambda t_k})=e^{-(n-k)\lambda t_k}$? I was solving one MLE(Maximum Likelihood Estimator) question and I encountered for this equation. How could I simplify $\prod_{j=k+1}^{n}(e^{-\lambda t_k})$ to the given answer? Because I'm unfamiliar with the $\prod$. Hope someone could help me out.
$$\prod_{j=k+1}^{n}(e^{-\lambda t_k})=e^{-(n-k)\lambda t_k}$$
 A: The $\prod_{j=k+1}^{n}$ in
$$\prod_{j=k+1}^{n}(e^{-\lambda t_k})=e^{-(n-k)\lambda t_k}$$
can be read as taking $n - k$ copies of the quantity after it, and multiplying it all together. Consider, for example, the case where $j = k + 1 = n$. Then it will simply be a single "copy". For $j = k + 1 = n - 1$, the "loop" will run for two iterations, etc.
Taking $n - k$ "copies" of that quantity behind it is equivalent to simply raising it to the power $(n - k)$. Since there already is a power up there, if you raise it to another power, you can simply multiply the powers. Doing that directly gives you the result.

Note that the above is only correct in your specific case because the "index" $j$ (the symbol at the bottom of the $\prod$) does not appear in the expression after the product symbol. If the $j$ would appear in that following expression, you would not multiply a sequence of "copies", but a sequence of expressions where you fill in varying $j$, starting with $j = k + 1$ and ending with $j = n$.
For example:
$$\prod_{j=1}^3 j = 1 \times 2 \times 3$$
or
$$\prod_{j=k+1}^n e^j = e^{k+1} \times e^{k+2} \times \dots \times e^{n}$$
A: Starting from $k+1$, how many steps are required to reach $n$? $k+1, k+2, ...,n.$
$$k+1(steps) = n$$
$$steps = n-k$$
As you aren't familiar with the $\prod$ its just a notation for multiplication of the terms. For e.g:
$$\prod_{j=1}^{3}x^j=x^1\cdot x^2\cdot x^3$$
Moreover, 
$$2^3 = 2\cdot 2\cdot 2 = \prod_{j=1}^{3}2$$
the term in the product is just repeated a number of times when it does not depend on the index.
Hence,
Let $C = e^{-\lambda t_k}$
$$\prod_{j=k+1}^{n}(C)=C^{n-k}$$
$$C = e^{-\lambda t_k}$$
$$C^{n-k} = ({e^{-\lambda t_k}})^{n-k} = e^{-(n-k)\lambda t_k}$$
A: $$\begin{align}
\prod_{j=k+1}^ne^{-\lambda{t_k}}
&=\overbrace{e^{-\lambda{t_k}}\cdot e^{-\lambda{t_k}}\cdot e^{-\lambda{t_k}}\cdots e^{-\lambda{t_k}}}^{n-(k+1)+1=n-k\text{ times}}\\
&=e^{-(\overbrace{\lambda t_k+\lambda t_k+\lambda t_k+\cdots+\lambda t_k}^{n-k\text{ times}})}\\
&=e^{-(n-k)\lambda t_k}\end{align}$$
