$\newcommand\la\lambda\newcommand\w{\mathfrak w}\newcommand\R{\mathbb R}$We have to show that $P(U<u)=u$ for $u\in(0,1)$, where 
$$U:=\min_{j\ge1} \frac{X_1+\cdots+X_j}j$$
and $X_1,X_2,\dots$ are iid exponential random variables with mean $1$. This minimum is attained almost surely (a.s.), because, by the strong law of large numbers, $\frac{X_1+\cdots+X_j}j\to1$ a.s. as $j\to\infty$, whereas $\inf_{j\ge1} \frac{X_1+\cdots+X_j}j<1$ a.s. 

For each natural $j$ and each $u\in(0,1)$, 
$$\begin{aligned}
U<u&\iff\exists j\ge1\ \;\sum_{i=1}^j X_i<ju \\ 
&\iff\exists j\ge1\ \;Y_{u,j}:=\sum_{i=1}^j(u-X_i)>0 \\ 
&\iff\bar Y_u>0, 
\end{aligned}\tag{1}$$
where $\bar Y_u:=\max_{j\ge0}Y_{u,j}$, with $Y_{u,0}=0$ (of course). 
By the formula $E e^{i\la\bar Y}=\w_+(\la)/\w_+(0)$ at the very end of Section 19 of Chapter 4 (p. 105) and Theorem 2 in this chapter (pp. 106--107) of [Borovkov][1], 
$$g_u(\la):=E e^{i\la\bar Y_u}=\frac{(1-u)i\la}{1+i\la-e^{i\la u}}$$
for all real $\la$. 
Note also that $\bar Y_u\ge Y_{u,0}=0$. 
So, by Proposition 1 in [this paper ][2] or its [arXiv version ][3], 
$$P(\bar Y_u>0)=E\,\text{sign}\,\bar Y_u
=\frac1{\pi i}\,\int_\R \frac{g_u(\la)}\la\,d\la
=\frac1{\pi i}\,\int_\R h_u(\la)\,d\la \tag{2}
,$$
where 
$$h_u(\la):=\frac{g_u(\la)-g_u(\infty-)}\la
=(1-u)\frac{1-e^{i \la u}}{\la(e^{i \la u}-1-i\la)}$$
and the integrals are understood in the principal value sense. 

In view of (1), it remains to show that the integrals in (2) equal $\pi i u$ for all $u\in(0,1)$. Numerical calculations suggest this is true. However, I have not been able to prove this, even with the help of Mathematica. 


  [1]: https://link.springer.com/book/10.1007/978-1-4612-9866-3
  [2]: https://www.sciencedirect.com/science/article/abs/pii/S0167715215002722?via%3Dihub
  [3]: https://arxiv.org/abs/1309.5928