Is there a simple algorithm for exponentiating large numbers to large powers?

I've been thinking about this for some days, a multiplication is a lot of sums, so:

$75\times 75=\stackrel{\text{75 times}}{\stackrel{\u23de}{75+75+75+75+75+75+75+75+\cdots}}$

But then, there is a simple algorithm that enable us to multiply without having to sum all those numbers. In the same way, exponentiation is repeated multiplication - but I wasn't taugth about such algorithm for exponentiation. I've been thinking in representing the number with a polynomial, for example:

$1038={10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0}$

Then:

${1038}^{1038}=({10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0}{)}^{1038}$

But from here (considering what I currently know) I'd have to multiply it $1038$ times. The mentioned multiplication would be:

$({10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0}{)}^{1038}=\stackrel{\text{1038 times}}{\stackrel{\u23de}{({10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0})\cdot ({10}^{3}+3\cdot {10}^{1}+8\cdot {10}^{0})\cdot \dots}}$

The first idea I had: There might be some connection with the binomial theorem, but I don't see how it fits. The second idea would be to find a way to write:

$({10}^{{3}}+3\cdot {10}^{{1}}+8\cdot {10}^{{0}}{)}^{1038}$

In which the red exponents are multiplied in some way by $1038$. There might be some connection with logarithms here, but I don't see it. And it could be the case that these techniques won't yield the results I'm looking for, so: Is there a simple algorithm for large numbers elevated to large exponents?