I know there is an operator there is this operator $$\sum_{i=0}^n i = \ 0+1+2+3+\dotsb+n$$ and there is $$\prod_{i=1}^n i = \ 1\cdot 2\cdot 3\cdot \dotsb \cdot n$$
But is there something like $\exp_{i=2}^5$ i that would give you $\ 2^{3^{4^5}}$?
I have search thought and everything I seen is the Knuth's up-arrow notation
If we introduce sum and product by the recursion $$\sum_{k=1}^0 a_k:=0,\qquad \sum_{k=1}^{n+1} a_k:=\sum_{k=1}^{n+1} a_k+a_{n+1},$$ $$\prod_{k=1}^0 a_k:=1,\qquad \prod_{k=1}^{n+1} a_k:=\prod_{k=1}^{n+1} a_k\cdot a_{n+1},$$ then it seems tempting to define in a completely analogoius fashion $$\mathop{\operatorname{EXP}}\limits_{k=1}^0a_k:=?,\qquad \mathop{\operatorname{EXP}}\limits_{k=1}^{n+1}a_k:=\left(\mathop{\operatorname{EXP}}\limits_{k=1}^{n}a_k\right)^{a_{n+1}}.$$ Unfortunately, this makes $$ \mathop{\operatorname{EXP}}\limits_{k=1}^{n}a_k=?^{\prod_{k=1}^na_k},$$ which is neither useful nor do we have a natural choice for $?$. The lack of commutativity in exponentiation plays tricks on us here, i.e., one has to be careful and define such an EXP operator with the recursion going backwards. So, you might rather want $$\mathop{\operatorname{EXP}}\limits_{k=1}^0a_k:=1,\qquad \mathop{\operatorname{EXP}}\limits_{k=1}^{n+1}a_k:= a_1^{\mathop{\operatorname{EXP}}\limits_{k=1}^{n}a_{k+1}}.$$ Yes, defining this is possible - we just did. However, the resulting operator seems to be of little use.