sequences and series - Prove that $limlimits_{n to infty} (1-frac{1}{2n+1})^{3n} = frac1{esqrt{e}}$

I have this problem that I cannot seem to solve. I tried splitting it into two factors $$\lim\limits_{n \to \infty} (1-\frac{1}{2n+1})^{2n}\times
\lim\limits_{n \to \infty} (1-\frac{1}{2n+1})^{n}$$ but that did not help because of the $+1$ in the denominator. So I tried multiplying both powers by $\frac{2n+1}{2n+1}$ but that did not help either. So now I'm stuck,and don't know what to do
can someone please help? Thanks

Answer

Hint: write $3n = \dfrac{3}{2}(2n+1) - \dfrac{3}{2}$, and use the fact that $\displaystyle \lim_{n\to \infty} \left(1-\dfrac{1}{2n+1}\right)^{2n+1} = \dfrac{1}{e}$. Can you take it from here?