We have by IIDness $P(Y_n\leq y)=\prod_{1\leq k \leq n}P(U_1\leq ky)=\prod_{1\leq k \leq n}(1\wedge ky),y \in [0,1]$. Now for all such $y$, there exists a $n(y)\in \mathbb{N}$ s.t. $yn> 1$ for all $n> n(y)$ so $\lim_nP(Y_n\leq y)=n(y)!y^{n(y)}$. In fact, $n(y)=\lfloor 1/y\rfloor$. So $\lim_nP(Y_n\leq y)=\lfloor 1/y\rfloor!y^{\lfloor 1/y\rfloor}$. To be sure that $Y_n$ actually converges in distribution to some rv, we set $\mathscr{F}_{n}=\sigma(U_k,k\leq n)$ and note: $$E[Y_n|\mathscr{F}_{n-1}]=E[\max(Y_{n-1},U_n/n)|\mathscr{F}_{n-1}]\geq Y_{n-1}$$ and $0\leq Y_n\leq 1$ for all $n$. So $(Y_n)_{n \in \mathbb{N}}$ is a bounded submartingale, so that it converges a.s. to some $Y$ by submartingale convergence. So $Y\sim \lfloor 1/y\rfloor!y^{\lfloor 1/y\rfloor}$ and $Y_n$ converges in distribution to $Y$.

We have by IIDness $P(Y_n\leq y)=\prod_{1\leq k \leq n}P(U_1\leq ky)=\prod_{1\leq k \leq n}(1\wedge ky),y \in [0,1]$. Now for all such $y>0$, there exists a $n(y)\in \mathbb{N}$ s.t. $yn> 1$ for all $n> n(y)$ so $\lim_nP(Y_n\leq y)=n(y)!y^{n(y)}$. In fact, $n(y)=\lfloor 1/y\rfloor$. So $\lim_nP(Y_n\leq y)=\lfloor 1/y\rfloor!y^{\lfloor 1/y\rfloor}$. To be sure that $Y_n$ actually converges in distribution to some rv, we set $\mathscr{F}_{n}=\sigma(U_k,k\leq n)$ and note: $$E[Y_n|\mathscr{F}_{n-1}]=E[\max(Y_{n-1},U_n/n)|\mathscr{F}_{n-1}]\geq Y_{n-1}$$ and $0\leq Y_n\leq 1$ for all $n$. So $(Y_n)_{n \in \mathbb{N}}$ is a bounded submartingale, so that it converges a.s. to some $Y$ by submartingale convergence. So $Y\sim \lfloor 1/y\rfloor!y^{\lfloor 1/y\rfloor}$ and $Y_n$ converges in distribution to $Y$.