How to calculate limit of the sequence $e^{-(n^frac{1}{2})}{(n+1)^{100}}$

Does the sequence $e^{-(n^\frac{1}{2})}{(n+1)^{100}}$ converge? If yes what is the limit?

What I tried: Expanding

$${(n+1)^{100}}= 1+^{100}C_1n+^{100}C_2{n^2}+^{100}C_3{n^3}+ \dots + ^{100}C_{100}{n^{100}}$$
Multiplying each term by $e^{-(n^\frac{1}{2})}$ and taking limits using L'Hospital Rule we get the limit of the sequence 0, but that is a lengthy and I think it is not a proper approach. [Actually applying L'Hospital rule twice to each term gives the preceding term(I checked it upto $[n^3e^{-(n^\frac{1}{2})}$] and hence ultimately limit will be 0 because lim($e^{-(n^\frac{1}{2})})=0$]
Can anyone please tell me whether this is a correct method to solve the problem and please suggest me a proper method if there is any. Thank you.

The answer is limit of the sequence is 0

Answer

Yes, the final limit is zero. Note that as $n\to +\infty$

$$e^{-\sqrt{n}}{(n+1)^{100}}=\exp\left({-\sqrt{n}\underbrace{\left(1-\frac{100\ln(n+1)}{\sqrt{n}}\right)}_{\to 1}}\right)\to0$$
because, for example by using L'Hopital,
$$\lim_{n\to +\infty}\frac{\ln(n+1)}{\sqrt{n}}=0.$$