Almost sure convergence and uniform integrability of a product of i.i.d random variables with density

I'm struggling with the following problem:

Let $$(\Omega, \mathcal{A}, \mathbb{P})$$ be a probability space, $$(X_n)_{n \in \mathbb{N}} \subseteq \mathcal{L}^0(\mathbb{P};\mathbb{R})$$ i.i.d with: $$\forall F \in \mathcal{B}(\mathbb{R}): \; \mathbb{P}(X_n \in F) = \int_F 1_{[1,e]} \frac{1}{z} \lambda^1(dz)$$ (Where $$1_{[1,e]}$$ is the indicator function). For all $$n \in \mathbb{N}$$ define $$Y_n := ( \prod\limits_{k=1}^n X_k )^{1/n}$$

a) Show that for all $$n \in \mathbb{N}$$ $$Y_n$$ is $$\mathcal{A}- \mathcal{B}(\mathbb{R})$$-measurable

I didn't have any issues solving this problem.

b) Show that $$(Y_n)_{n \in \mathbb{N}}$$ converges almost surely against 1.648721....

I tried using the law of large numbers. I calculated $$\log((Y_n)^n) = \sum\limits_{k=1}^n \log(X_k)$$ The sequence $$(\log(X_k))_{k \in \mathbb{N}}$$ is still i.i.d. , thus for $$k \in \mathbb{N}$$ I tried to calculate the expected value $$\mathbb{E}[\log(X_k)] = \int_\mathbb{R} \log\left(1_{[1,e]} \frac{1}{z}\right) \lambda^1(dz)$$ But when I tried to calculate this integral, it is undefined. Where did I go wrong? Another approach I thought about was trying to use the almost completely convergence, as in $$\sum\limits_{n=1}^\infty \mathbb{P}( |Y_n - Y| > \epsilon) < \infty$$ for any $$\epsilon \in (0,\infty)$$, in order to imply almost sure convergence. I thought about this idea because the limit is given incompletely as a decimal representation, and not as (most likely) $$\sqrt{e}$$. But I can't make any significant progress with that idea

c) Show that $$(Y_n)_{n \in \mathbb{N}}$$ is uniformly integrable.

I tried using $$L^1$$-convergence to conclude that $$(Y_n)_{n \in \mathbb{N}}$$ is uniformly integrable. But here I encounter the same problem as in b), that I can't seem to solve the integral at hand and can't calculate the expectation.

I would be very thankful for any advice or progress to the solution!

Actually, $$\mathbb{E}[\log(X_k)] = \int_\mathbb{R}1_{[1,e]} \log\left(z\right)\frac 1z \lambda^1(dz)$$ and the integrand is the primitive of $$(\log z)^2/2$$ hence $$\mathbb{E}[\log(X_k)]=1/2$$. Then the law of large numbers show that $$\log Y_n\to 1/2$$ almost surely.
For uniform integrability, it is a good idea to show convergence in $$\mathbb L^1$$. Since $$Y_n\to \sqrt e$$ almost surely and $$Y_n$$ is non-negative, it suffices to show that $$\mathbb E[Y_n]\to \sqrt e$$. As $$\mathbb E[Y_n]$$ can be explicitely computed (indeed, $$\mathbb E[Y_n]=\left(n\left(e^{1/n}-1\right)\right)^n$$, this becomes now a calculus problem.
Or in a simpler way: $$1\leqslant X_i\leqslant e$$ for each $$i$$ hence $$1\leqslant Y_n\leqslant e$$.