calculus - Help solving the limit $lim_{nto infty} left(1+frac{1}{n!}right)^{2n}$

$$\lim_{n\to \infty} \left(1+\frac{1}{n!}\right)^{2n} $$

I've tried:

$$\lim_{n\to \infty} \left(1+\frac{1}{n!}\right)^{2n} = \lim_{n\to \infty} e^{\ln{\left(1+\frac{1}{n!}\right)^{2n}}} = \lim_{n\to \infty} e^{2n\ \ln{\left(1+\frac{1}{n!}\right)}}$$
But I don't know how to work with the factorial

Answer

$$\lim_{n\to \infty} \left(1+\frac{1}{n!}\right)^{2n}=\lim_{n\to \infty}\left( \left(1+\frac{1}{n!}\right)^{n!}\right)^{\frac2{(n-1)!}}=e^0=1$$

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Alternatively,
Let $$A=\left(1+\frac{1}{n!}\right)^{2n}$$

So, $$\log A=2n\log \left(1+\frac{1}{n!}\right)\text { using }\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots$$

which comes from the Taylor series which holds for $|x|<1$ as $n\to\infty, \frac1{n!}\to0\implies |\lim_{n\to\infty}\frac1{n!}|<1 $

$$=2n\left(\frac1{n!}-\frac{\frac1{(n!)^2}}2+\frac{\frac1{(n!)^3}}3+\cdots\right)$$

$$=\frac2{(n-1)!}-\frac1{n!(n-1)!}+\frac{2}{3(n-1)!(n!)^2}+\cdots$$

So, $$\lim_{n\to \infty}\log A=0\implies \lim_{n\to \infty}A=e^0=1 $$