Find the distribution of $ N = \min \left\{k: \prod_{i = 1}^{k}U_i \lt .6\right\}. $ I'm cross-posting this from math.SE because it's not getting any love over there. However, if that's considered heresy, I can delete the posting over there.
The Statement of the Problem:
Let $ \{ U_i \}$ be a set (sequence?) of iid random variables such that $U_i \sim \text{Uniform}(0,1)$, and define
$$ N(a) = \min \left\{k: \prod_{i = 1}^{k}U_i \lt .6\right\}. $$
Find the distribution of $N$.
SOLUTION
Recall that $-\log (U) \sim \text{Expo}(1)$. From this, it follows that
$$N(a) = \min \left\{k: \prod_{i = 1}^{k}U_i \lt .6\right\} = \min \left\{k: \sum_{i = 1}^{k} \left[ - \log ( U_i ) \right] \gt \log \left( \frac{5}{3} \right) \right\} = \min \left\{k: S_k \gt \log \left( \frac{5}{3} \right) \right\} $$
where $S_k$ is the arrival time of the $k$th event of a Poisson process with rate $\lambda = 1$. I understand all of this. Beyond this point is where I get confused...
Now, it follows (apparently) that the smallest $k$ such that $S_k \gt \log \left( \frac{5}{3} \right)$ is 
$$ N\left( \log \left[ \frac{5}{3} \right] \right) + 1 \qquad (*) $$
which is distributed as $\text{Poisson}{ \left( \log \frac{5}{3} \right) } + 1$.
I'm not sure what is meant by this sort of "superscript" parameter here (I'm transcribing this from handwritten notes). As far as I can tell, it's just a normal ol' parameter. Anyway, I have no idea where the result $(*)$ comes from, and have been racking my brain here trying to figure it out. Any assistance would be appreciated.
 A: Takings logs both sides is smart. Here, just for fun ... is a more brute force approach ...
Let $X_i \sim \text{Uniform}(0,1)$ be iid, and let $Y = \prod_{i = 1}^{k}X_i$ denote the product of $k$ independent Uniforms. Then, the pdf of $Y$, say $h(y)$, can be shown (I used method of induction) to be:

We seek $P(Y<.6)$. More generally, for parameter $a$, the $P(Y<a)$ is:

where I am using the Prob function from the mathStatica package for Mathematica to automate the nitty-gritties, and Gamma[k,z] is the incomplete gamma function $\Gamma(k,z)=\int _z^{\infty } t^{k-1} e^{-t} d t$.
Here is a plot of the solution $P(Y<.6)$ as a function of $k$:

Given cdf sol above, we can derive the corresponding pmf (in $k$) by evaluating $\text{sol}(k) - \text{sol}(k-1)$, which yields the pmf $f(k)$:
$$f(k) = \frac{a}{(k-1)!} (-\log (a))^{k-1} \quad \text{for } k = 1,2,3, \dots$$
Note that this does indeed have the Poisson form $\frac{e^{-\lambda } \lambda ^x}{x!}$ where $\lambda = -\log (a)$, and $x = k-1$.
Here is a plot of the corresponding pmf $f(k)$ (again when $a = .6$):

