This is an answer to the request in some comments to show how to arrive at the formula for the entries of $L^h$
Here is an example how we can find the entries of $L^h$ symbolically. I do with the $4 \times 4 $ triangular matrix $L$ $ M(h) = \exp( h \cdot \log (L)) = L^h $ and get for $M$ $ M(h) = \left[ \begin{array} {rrrr} 1 & . & . & . \\ 1h & 1 & . & . \\ \frac 12( 1 h^2+ 1 h) & 1h & 1 & . \\ \frac 16( 1h^3+ 3 h^2+ 2 h) & \frac 12 (1h^2+ 1h) & 1h & 1 \end{array} \right]$ Here a trained eye recognizes the Stirling numbers 1st kind as coefficients at the powers of $h$ and because the structure of the matrix has this constant diagonals it is easy to make the formula for the transfer: $ M(h)\cdot S_k = S_{k+h} $ One more step shows, that the evaluation of the polynomials in the entries leads to binomial numbers, which is a well known property of the Stirling numbers first kind (the vandermonde-matrix LDU-decomposes into the matrices of Stirling-numbers 2st kind and of binomial coefficients and thus reduces by the multiplication with the matrix of Stirling numbers 1'st kind (which is the inverse of the 2nd-kind matrix) to binomials)
I had fun to proceed a bit. Factorizing the smbolic entries, assuming the hypothese that we have always the Stirling numbers 1st kind and the fractional cofactors the reciprocal fatorials give
$ M(h) = \left[ \begin{array} {llll} 1 & . & . & . \\ 1(h) & 1 & . & . \\ \frac 12( h(h+1)) & 1(h) & 1 & . \\ \frac 16( h(h+1)(h+2)) & \frac 12 ( h(h+1)) & 1(h) & 1 \end{array} \right]$ and this gives immediately the binomial coefficients
$ M(h) = \left[ \begin{array} {cccc} 1 & . & . & . \\ \binom{h}{1} & 1 & . & . \\ \binom{h+1}{2} & \binom{h}{1} & 1 & . \\ \binom{h+2}{3} & \binom{h+1}{2} & \binom{h}{1} & 1 \end{array} \right]$ and a routine could solve the problem given the vector S of dimension n in the following way:
T = 0*S ; \\ initialize an empty array of size of S b = 1; \\ contains the current binomial for(j=1,n, for(k=j,n, T[k]+=S[k+1-j]*b); b *= (h+j-1)/j; ); return(T);
So we have $n^2/2$ operations by the looping.