Speed of the General number field sieve So according to the wikipedia page https://en.wikipedia.org/wiki/General_number_field_sieve the algorithm has complexity $$\exp \left( \left(\sqrt[\leftroot{1}\uproot{0}3]{\frac{64}{9}} + o(1) \right) (\ln n)^{\frac{1}{3}}(\ln \ln n)^{\frac{2}{3}} \right)$$ but I am not too familiar with this notation.

I understand the big-O notation for complexity if somebody could explain how this is similar and what it means for the running time it would be appreciated!!

Also I know that this is subexponential in the size of $n$ running time, but I am unsure how to show this just from this complexity.
Since the numeric constant in the exponent is approximately $1.923 $ the expression for the average number of iterations, say time complexity $T$  given by
$$T=\exp \left( \left(\sqrt[\leftroot{1}\uproot{0}3]{\frac{64}{9}} + o(1) \right) (\ln n)^{\frac{1}{3}}(\ln \ln n)^{\frac{2}{3}} \right)$$ 
is proportional to (for $n$ large enough so that $o(1)$ term is negligible)
$$T\propto \exp \left(1.923 \times  (\ln n)^{\frac{1}{3}}(\ln \ln n)^{\frac{2}{3}} 
\right)$$ 
which is upper bounded as (ignoring a multiplicative constant in front)
$$T\leq \exp \left(2   (\ln n)^{\frac{1}{3}}(\ln \ln n)^{\frac{2}{3}}  \right)=T_0^2.$$ 

Clearly $T$ is polynomial/exponential if and only if $T_0$ is.

Now consider the more general expression
$$f(\alpha,n)=\exp \left(  (\ln n)^{\alpha}(\ln \ln n)^{1-\alpha} 
 \right),\quad \alpha \in [0,1],$$
and note that when $\alpha=1,$ $f(1,n)=\exp(\ln n)=n,$ hence it is exponential in the bitsize $\log_2 n$ of the input $n.$

Also note that when $\alpha=0,$ $f(0,n)=\exp(\ln\ln n)=\ln n,$ hence it is polynomial in the bitsize $\log_2 n$ of the input $n.$

However we have $T_0=f(1/3,n)$ with $\alpha=1/3$ above so the complexity we have is a linear combination in the exponent of polynomial and exponential complexities. This is referred to as subexponential but superpolynomial complexity.