Analytically finding the value of $n$ for which $n \prod\limits_{k=0}^{n-1}\frac{60-k}{60}$ is maximized I was solving a problem where I needed to find the value of $n\in [0,60] \cap \mathbb N$ for which $$f(n)=n \prod_{k=0}^{n-1}\frac{60-k}{60}$$ is maximized.
Somewhat irresponsibly, I computed the value of the expression for $n=1,2,...,9$, saw that it decreases after $n=8$ and assumed that $n=8$ gives the maximum value.
Not only does this assume that the value of $f$ "increases then decreases," allowing only one candidate for a maximum, but it also doesn't shed any light on how I would solve the question or graph the function in the general case.
Therefore, I have 3 questions:
(1) How would I find the maximum (or maxima, if unsure of how many extrema there would be) analytically?
(2) Is it trivial to show that $f$ "increases then decreases"?
(3)  How might I sketch $f$ (by hand) ?
Can the answers for the above questions be generalized for the general case function $g(n)=n \prod_{k=0}^{n-1}\frac{\alpha-k}{\alpha}$ for $n\in [0,\alpha] \cap \mathbb N$?
 A: Partial answer, points (1) and (2).
$$g(n+1)=(n+1)\prod\limits_{k=0}^{n}\frac{\alpha-k}{\alpha}=
n\prod\limits_{k=0}^{n}\frac{\alpha-k}{\alpha} + \prod\limits_{k=0}^{n}\frac{\alpha-k}{\alpha}=\\
g(n)\frac{\alpha-n}{\alpha}+\frac{g(n)}{n}\frac{\alpha-n}{\alpha}=\\
g(n)\frac{\alpha-n}{\alpha}\frac{n+1}{n}$$
To have a growth, we ask for
$$\frac{\alpha-n}{\alpha}\frac{n+1}{n}\geq1 \tag{1}$$
because that will make $\color{red}{g(n+1)\geq g(n)}$. This happens for:
$$(\alpha-n)(n+1)\geq \alpha n \Leftrightarrow \\
\alpha n+ \alpha - n^2 -n \geq \alpha n \Leftrightarrow\\
n^2 +n - \alpha \leq 0$$
or
$$n \leq \frac{-1+\sqrt{1+4\alpha}}{2} \tag{2}$$
As a result, the growth finishes at around $\left \lfloor  \frac{-1+\sqrt{1+4\alpha}}{2} \right \rfloor$ or $\left \lfloor  \frac{-1+\sqrt{1+4\alpha}}{2} \right \rfloor +1$. For $n >\left \lfloor  \frac{-1+\sqrt{1+4\alpha}}{2} \right \rfloor +1$ we have
$$\frac{\alpha-n}{\alpha}\frac{n+1}{n}<1 \tag{3}$$
which means $\color{red}{g(n+1)<g(n)}$ and the sequence will start decreasing, thus the maximum is reached at around $\left \lfloor  \frac{-1+\sqrt{1+4\alpha}}{2} \right \rfloor$ or $\left \lfloor  \frac{-1+\sqrt{1+4\alpha}}{2} \right \rfloor +1$
For $\alpha=60$ we have, from $(2)$, that $n\approx 7$ or $8$, which can be seen here. For $\alpha=160 \Rightarrow n\approx 12$ or $13$ 
