Knowing that the number $e =\lim\_{n\to\infty}\left(1+{1\over{n}}\right)^n$ solve $a\_n=\left({n+1}\over{n+3}\right)^n$

So (...)
$$\lim\_{n\to\infty}\left({n+1}\over{n+3}\right)^n = \lim\_{n\to\infty}\left({n+1+2-2}\over{n+3}\right)^n = \lim\_{n\to\infty}\left(1-{2\over{n+3}}\right)^n = \lim\_{n\to\infty}\left({1+{{1\over{-n-3}}\over2}}\right)^n = \lim\_{n\to\infty}\left({1+{{1\over{-n-3}}\over2}}\right)^{{-n-3\over2}.{2\over{-n-3}}.n} = e^{\lim\_{n\to\infty}{2n\over{-n-3}}}=e^{-2}$$

I know to solve it in this way, but my teacher told me that the in exams I can NOT solve it in this way. 
The problem is that I can not put the denominator in the exponent and then multiply by the inverse to make it 1 (note that this does not alter the exercise).

Is there another way to do this?