calculus - $ text{Prove:} , {a_n}>0, displaystyle{lim_{n to infty}} { a_n }=L Rightarrow displaystyle{ lim_{n to infty } {a_n }^frac{1}{n}}=1 $

I have this exercise to prove in calculus 1:

$$ \text{Prove:} \, \{a_n\}>0, \displaystyle{\lim_{n \to \infty}} \{ a_n \}=L \Rightarrow \displaystyle{ \lim_{n \to \infty } \{a_n \}^\frac{1}{n}}=1 $$

Where "$\{a_n\}$" is a sequence.

We proved in class that $ \displaystyle{\lim_{{n \to \infty }} \cfrac{1}{n}} = 0 $.

Any $L^0=1$.

$\displaystyle{\lim_{n \to \infty}} \{ a_n \}=L,L^0=1, \displaystyle{\lim_{n \to \infty}} \frac{1}{n} = 0 \Rightarrow \\ \displaystyle{\lim_{n \to \infty}} \{ a_n \} ^\frac{1}{n} = L^0=1 \\ QED$

Is my solution wrong? The solution of the university is different.
Their solution is:

$\text{Because} \, a_n \to L \, \text{then for} \, \epsilon = \frac{L}{2} \, \text{exists} \, N \, \text{such that} \, \forall n>N: \\ \cfrac{L}{2} = L-\cfrac{L}{2}