Anne has the winning strategy. At first turn she replace $10^{2007}$ with $2^{2007}$ and $5^{2007}$. Note that $2$ and $5$ are prime therefore on the board will be only powers of $2$ and $5$. Let $X$ be the set of powers of $2$ written on the board and $Y$ be the set of powers of $5$ written on the board. Note that during the turn one can only change numbers from one of the sets $X,Y$ because $X\cap Y=\emptyset$. Now whenever Berit changes set $X$ i.e. $Y$ Anne can do the same thing with the set $Y$ i.e. $X$ (if she switch $2$ and $5$), therefore she will win since game must be finished (product of all numbers never groves and one can do 1) only finite times).