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On page 124 of Introduction to Set Theory, Jech claims that ordinal functions below are continuous in the second variable: If $gamma$ is a limit ordinal and $beta= sup_{v<gamma} beta_{v}$ then
$$alpha + beta = sup_{v<gamma} (alpha + beta_{v})$$
$$alpha . beta = sup_{v<gamma} (alpha . beta_{v})$$
$$alpha ^ beta = sup_{v<gamma} (alpha ^ {beta_{v}})$$
Then it is said that the results above are followed directly from the definitions!

Somebody has answered something over here which somehow asks the same question but the problem is question above which is directly taken from Jech book is stated in a slightly different way considering indexes and it has made some confusion for me since I am new to the topic. I would appreciate if you could clarify the exact arguments above.

I am going to explain how to prove the continuity of addition, the rest facts can be proved in similar fashion. First of all, we should keep in mind the monotonicity of addition in the second variable. We now that $$alpha + beta = sup{alpha + rhocolon rho<beta}.$$
WLOG we can consider $beta$ to be a limit ordinal. Hence for every $nu<gamma$ we have $beta_nu < beta$, hence $alpha + beta_nu < alpha + beta$, hence by the definition of supremum $$sup{alpha + beta_nucolon nu<gamma} leq alpha + beta.$$
It remains to prove the converse inequality. Let $rho<beta$. There exists $nu_0<gamma$ such that $rho<beta_{nu_0}$ (otherwise all $beta_nu$ would be less or equal than $rho<beta$). By monotonicity $alpha + rho < alpha + beta_{nu_0}$, so $alpha+rho<sup{alpha+beta_nucolon nu<gamma}$, again by the definition of supremum we get
$$
alpha+beta leq sup{alpha+beta_nucolon nu<gamma}.
$$
These two inequalities give us the desired equality.